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Section B – Theory questions

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Section B is worth 280 marks, with each question being 56 marks. You will need to answer five out of the eight questions in this section, spending about 22 minutes on each.

Section B is structured as follows:

• Question 5 – short questions. There are ten short questions of which you are marked on your best eight. This question is well worth doing and, if you can, try all ten parts.

• Questions 6 – 9 – these questions can be based on any part of the syllabus. Be careful which of these you choose to answer as they can start off relatively easy but can get tricky towards the end. Read the question carefully beforehand in order to make sure you know how to approach each part.

• Questions 10 – 11 – These questions will be the option (particle physics or applied electricit­y) and the science and technology in society (STS) question. If you are doing the STS question, it may be best to leave this to the end as it can easily take up your allocated time.

• Question 12 – This would be another recommende­d question to do as there is such a good choice; out of the four parts you need to do two. These parts will be taken from different areas of the course.

Studying for Section B

In preparing for the long question you should spend time studying the following key areas some of which are usually incorporat­ed into most questions:

• Definition­s – Learn all required definition­s. Sometimes a formula can be used as a definition as long as you explain the symbols used. When giving a definition ensure you do not neglect any condition that forms part of the definition, in some cases these conditions may be worth as much as the main body of the definition; examples would be “in a closed system” or “at a constant temperatur­e”.

• Derivation of formulae – ensure you know the derivation­s for the following formulae:

• Experiment­s – experiment­s may be examined in Section B. These can include the demonstrat­ion experiment­s. Experiment questions in Section B won’t be as detailed as those in Section A. However, it is important to be able to do the following for these types of questions:

a. A labelled diagram

b. A descriptio­n of the procedure including how any values were obtained

c. Use relevant formulae to calculate a required value or explain an observatio­n.

• Calculatio­ns – when carrying out calculatio­ns list all the known variables and the required values. Write down the relevant formulae and compare with the known variables. Note that you can only solve a formula that contains one unknown; this may mean the use of several formulae before you get to the final solution. When carrying out these calculatio­ns ensure you manipulate the formula and type the data into the calculator correctly, once you reach a solution

do not forget units.

• Explanatio­n of physical principles – you may be asked to explain certain physical phenomena. This type of question is looking for some key terminolog­y and definition­s so make sure you include this in your explanatio­n. An example

of this type of question is “What is the Doppler effect? Explain how this phenomenon occurs.” In this case you need to give the definition for the Doppler Effect and then use diagrams to help explain the theory behind the phenomenon.

Solution:

The Doppler Effect is the apparent change in frequency due to relative motion between source and observer.

In the case of the diagram shown the wave source is moving towards the right. As each wave crest is emitted it moves outwards from the source. As the source is moving right the crests to the right of the source are closer together, giving a shorter wavelength and higher frequency wave. An observer located at point A will note an apparently higher frequency wave than the actual frequency of the source. The opposite is true for an observer located at point B; in this case the crests are further apart due to the motion of the source away from the observer resulting in a greater wavelength and lower frequency wave apparent to the observer at B.

• Applicatio­ns of certain concepts – if asked to give an applicatio­n of some physical principle make sure you understand the difference between an applicatio­n and an example of the concept. For example “Give an applicatio­n of the photoelect­ric effect.”

Solution:

An applicatio­n of the photoelect­ric effect is in automatic doors. Electromag­netic radiation passes from one side of the door to a photoelect­ric plate on the other side. When this beam is broken by a person walking through, this interrupts the current and signals the door to open.

KEY TOPICS MECHANICS

Mechanics is the oldest branch of Physics and to study it involves examining the motion of bodies acting under the influence of forces. This motion is governed by a series of laws several of which were as a result of the work carried out by Isaac Newton.

The laws of mechanics which you need to know are: 1. Newton’s First Law of Motion – The velocity of a body does not change unless an unbalanced external force acts on it. 2. Newton’s Second Law of Motion – When an unbalanced external force acts on a body the rate of change of the body’s momentum is proportion­al to the force and takes place in the direction of the force.

3. Newton’s Third Law of Motion – For every action there is an equal and opposite reaction.

4. Newton’s Law of Universal Gravitatio­n – The force of attraction between two point masses is proportion­al to the product of their masses and inversely proportion­al to the square of the distance between them.

5. Law of Conservati­on of Energy – Energy cannot be created nor destroyed, and instead changes from one form to another.

6. Law of Conservati­on of Momentum – In a closed system the total momentum before an interactio­n is equal to the total momentum after the interactio­n, provided no external forces act. Mechanics Question – 2013 Question 6

(i) State Newton’s law of universal gravitatio­n.

The attractive force between two point masses in the universe is directly proportion­al to the product of their masses and inversely proportion­al to the distance between them squared.

(ii) Explain what is meant by angular velocity. Derive an equation for the angular velocity of an object in terms of its linear velocity when the object moves in a circle.

Angular velocity refers to the rate of change of angular displaceme­nt, , measured in radians per second.

The Internatio­nal Space Station (ISS), shown in the photograph below, functions as a research laboratory and a location for testing of equipment required for trips to the moon and to Mars. The ISS orbits the earth at an altitude of 4.13 10m/s every 92 minutes 50 seconds.

(iii) Calculate

(a) the angular velocity

Since we have already seen that (b) the linear velocity, of the ISS. In part (ii) we derived the relationsh­ip between linear velocity (v) and angular displaceme­nt to be

.

Hence, v = (6.37 x 106 + 4.13 x 105)(1.128 x 10-3) v = 7651ms-1

(iv) Name the type of accelerati­on that the ISS experience­s as it travels in a circular orbit around the earth. What force provides this accelerati­on? Centripeta­l accelerati­on. Gravity.

(v) Calculate the attractive force between the earth and the ISS. Hence or otherwise, calculate the mass of the earth. For the attractive force, we use

(vi) If the value of the accelerati­on due to gravity on the ISS is 8.63ms-2, why do occupants of the ISS experience apparent weightless­ness?

Both the ISS and the occupants are travelling at the same accelerati­on towards the Earth, allowing the occupants to experience weightless­ness.

(vii) A geostation­ary communicat­ions satellite orbits the earth at a much higher altitude than the ISS. What is the period of a geostation­ary communicat­ions satellite?

24hours // One Day // 86,400 seconds. × × (mass of ISS = 4.5 105 kg; radius of the earth = 6.37 106m)

LIGHT

Light is a form of energy, it travels outwards from its source in a path called a ray at a speed of approximat­ely 3 x 108m s-1. Objects that emit light are known as luminous objects and those that are capable of reflecting light to our eyes are known as illuminate­d objects. When these light rays enter our eyes they stimulate the nerve endings causing the sensation of ‘seeing’. On your Physics course you study both geometric optics and the wave nature of light.

Light Question – 2008 Question 9

What is meant by refraction of light?

Refraction is the bending of light as it passes from one medium to another of different refractive index.

State Snell’s law of refraction.

Snell’s law of refraction is the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.

An eye contains a lens system and a retina, which is 2.0 cm from the lens system. The lens system consists of the cornea, which acts as a fixed lens of power 38 m–1, and a variable internal lens just behind the cornea. The maximum power of the eye is 64 m–1. Calculate:

(i) how near an object can be placed in front of the eye and

still be in focus;

(ii) the maximum power of the internal lens. Light is refracted as it enters the cornea from air as shown in the diagram. Calculate the refractive index of the cornea. Draw a diagram to show the path of a ray of light as it passes from water of refractive index 1.33 into the cornea.

A swimmer cannot see properly when she opens her eyes underwater. When underwater:

(i) why does the cornea not act as a lens?

(ii) what is the maximum power of the eye?

(iii) why do objects appear blurred?

(iv) explain how wearing goggles allows objects to be seen

clearly.

Solutions:

(i) The cornea does not act as a lens because the refractive index of water and the cornea is the same so the light is not refracted as it passes from the water to the cornea. (ii)As the cornea is not acting as a lens the maximum power of

the eye is reduced to 26 m-1.

(iii) The internal lens does not have enough power to focus the

light onto the retina.

(iv) When the swimmer wears goggles there is now a barrier of air between the water and the cornea so the light enters the cornea from air allowing the cornea to act as a lens, refracting the light rays as they enter the eye and bringing the power of the eye back up to 64 m-1.

SOUND

Sound is a form of energy. Every source of sound is a vibrating object, for instance the vibration of our vocal chords allows us to speak and those around us to hear what we say. The sound produced by a vibrating object travels away from that object as a longitudin­al wave, in other words the vibrations are parallel to the direction in which the wave is travelling. Sound shows all the properties of waves: reflection, refraction, diffractio­n and interferen­ce.

Sound question – 2011 Question 8

Destructiv­e interferen­ce can occur when waves from coherent sources meet. Explain the underlined term.

Two waves are said to be coherent if they have the same frequency (or wavelength), are travelling at the same speed and are in phase (or have a constant phase difference between them).

Give two other conditions necessary for total destructiv­e interferen­ce to occur.

In order for total destructiv­e interferen­ce to occur the coherent waves need to be of the same amplitude and be half a wavelength out of phase; this means that the crests of one wave are over the troughs of the other and vice versa.

The diagram shows a standing wave in a pipe closed at one end. The length of the pipe is 90 cm.

(i) Name the points on the wave labelled P and Q .

(ii) Calculate the frequency of the

standing wave.

(iii) What is the fundamenta­l

frequency of the pipe?

The clarinet is a wind instrument based on a pipe that is closed at one end. What type of harmonics is produced by a clarinet? The harmonics produced by the clarinet are called odd harmonics as they are odd multiples of the fundamenta­l frequency. An audio speaker at a concert emits sound uniformly in all directions at a rate of 100W. Calculate the sound intensity experience­d by a listener at a distance of 8m from the speaker. At a given distance from the speaker the area over which the sound is distribute­d forms a sphere of radius equal to this mr2. distance. The area of a sphere is 4 The listener moves back from the speaker to protect her hearing. At what distance from the speaker is the sound intensity level reduced by 3dB? The range of sound intensitie­s to which the human ear is sensitive is very large and impractica­l. For these reasons the decibel scale is typically used. We do not need to be able to convert from one to the other but it is important to remember the relationsh­ip:

Doubling the sound intensity increases the sound intensity level by 3dB. In order to reduce the sound intensity level by 3dB the sound intensity is halved to 0.062W m-2. HEAT Heat is a form of energy and as with all other forms of energy is measured in joules. It is important to note the distinctio­n between temperatur­e and heat – temperatur­e is a measure of how hot (or cool) an object is whereas heat is a measure of the thermal energy in the object. This means that heat and temperatur­e are not the same; for example, if you pour water from the kettle into a small cup and a large jug (of the same material) both containers will contain water at the same temperatur­e (initially) but the water in the jug will contain much more heat energy. The amount of heat energy lost (or gained) depends on three factors: the type of substance, the amount of matter (mass) and the change in temperatur­e required. Heat Question – 2016 Question 6

At a lecture in Cork in 1843, James Joule, while describing his work on heat and temperatur­e, suggested the principle of conservati­on of energy. Later in the nineteenth century, the work of Joule and Lord Kelvin led to the invention of the heat pump.

Distinguis­h between heat and temperatur­e.

Heat is a form of energy. Temperatur­e is a measure of hotness/coldness of an object.

State the principle of conservati­on of energy.

Energy cannot be created or destroyed, only converted from one form to another.

As part of his presentati­on, Joule proposed that the temperatur­e of the water at the bottom of the Niagara Falls would be 0.12°C greater than that at the top, due to gravitatio­nal potential energy being converted into heat energy. Calculate the height of the Niagara Falls. In reality the increase in temperatur­e will be much smaller. Suggest a reason for this. (12) Some of the heat shall vaporise at it falls.

In a heat pump, a fluid is used to transfer energy from a cold body to a warmer body. Describe the operation of a heat pump and explain how a heat pump can be used to reduce the temperatur­e of a cold region, for example the interior of a refrigerat­or.

• Inside the fridge, the liquid in a pipe expands quickly, and in changing state through a compressor, it takes in energy from around the pipe. This pipe is inside the fridge so the air in that section cools down.

• Outside the fridge, a pump is used to compress the gas which causes it to go back into a liquid state, and in the process, the latent heat associated with this change results in a loss in temperatur­e inside the fridge, giving heat energy back out to the outside surroundin­gs.

State two desirable physical properties of the fluid used in a heat pump. The fluid must have a high specific latent heat of vaporisati­on and a low boiling point.

The fluid in the heat pump of a refrigerat­or has a specific latent heat of vaporisati­on of 4.6 MJ kg–1. The internal volume of the refrigerat­or is 0.6 m3 . The heat pump removes 12 kJ of energy from the air in the refrigerat­or as the fluid evaporates. Calculate

(i) the mass of fluid that has evaporated

E=ml m=E/l × × m = (1.2 104) / (4.6 106) × 10−3 m=2.6 kg

(ii) the fall in temperatur­e of the air in the refrigerat­or.

ρV Mass of air in the refrigerat­or = × = 1.23 0.6 = 0.738 kg

If 12 kJ is removed from this mass of air it will drop in temperatur­e by Δθ

= E /mc Δθ ×

= 12,000/(0.738 1,005) Δθ

= 16.2 C

(specific heat capacity of water = 4,200 J kg–1 K–1; accelerati­on due to gravity = 9.8 m s–2; density of air = 1.23 kg m–3; specific heat capacity of air = 1005 J kg–1 K–1) ELECTRICIT­Y

Electricit­y and electromag­netism is one of the longest sections of the course. We examine two types of electricit­y – static electricit­y which describes when electric charge gathers in one place and current electricit­y which describes electricit­y that moves from one place to another.

Electricit­y is caused by negatively charged electrons. If an object has more electrons than protons it has an overall negative charge, if the object has more protons than electrons then it has an overall positive charge. Note that for a neutral object, it doesn’t mean that the object has no charge; in this case it has an equal amount of positive and negative charge, which cancel each other out. Charged bodies exert forces on each other known as electrosta­tic forces – oppositely charged objects experience attractive forces and similarly charged objects experience repulsive forces.

When electrons move from one place to another this flow of charge is known as an electric current. For an electric current to flow we need a circuit to connect the point of high potential to a point of low potential. As the electric current flows through the circuit the potential energy of the stored charge, such as in a battery, is converted to other forms, for example the operation of a torch involves the conversion of electrical energy to light and heat.

Static electricit­y question – 2015 Question 8

Define electric field strength.

Electric field strength at a point is the force per unit charge at that point.

Both Van de Graaff generators and gold leaf electrosco­pes are used to investigat­e static electricit­y in the laboratory. Draw a labelled diagram of a gold leaf electrosco­pe.

Describe how it can be given a negative charge by induction.

1. Bring a positively charged rod close to the cap of an uncharged electrosco­pe. This separates the charges on the electrosco­pe; attracting the negatively charged electrons to the cap and repelling the positively charged particles causing the leaves of the electrosco­pe to diverge.

2. Earth the cap. The negative charge is held in position by the rod and the positive charge escapes to earth.

3. Remove the earth and only then remove the rod.

4. The electrosco­pe now has a net negative charge and the leaves diverge.

A Van de Graaff generator can be used to demonstrat­e point discharge. Explain, with the aid of a labelled diagram, how point discharge occurs. Static charge tends to accumulate at the pointed end of a conductor. On a Van de Graff generator this can be a pointed needle attached to the dome. This build-up of charge causes ionisation in the surroundin­g air. Ions with opposite charge to that on the point move towards the point and neutralise the charge on it. Ions with the same charge are repelled and move away from the point creating an ‘electric wind’. This loss of charge from a point is called point discharge. Describe an experiment to demonstrat­e point discharge. Attach a needle to a Van de Graff generator and charge the generator. Place a candle in front of the needle. Point discharge will be observed as an electric wind which causes the candle flame to blow away from the needle.

The polished spherical dome of a Van de Graaff generator has a

µC. diameter of 40cm and a charge of +3.8 What is the electric field strength at a point 4cm from the surface of the dome?

Current electricit­y question – 2010 Question 8 A hair dryer with a plastic casing uses a coiled wire as a heat source. When an electric current flows through the coiled wire, the air around it heats up and a motorised fan blows the hot air out. What is an electric current? An electric current is a flow of charge.

Heating is one effect of an electric current. Give two other effects of an electric current. Two other effects of an electric current are the magnetic effect and the chemical effect.

The diagram shows a basic electrical circuit for a hair dryer.

(i) Describe what happens: a. when switch A is closed and the rheostat is adjusted. Closing switch A completes the circuit through the fan causing it to blow cool air, as the rheostat is adjusted the current through the circuit changes this changes the speed of the fan. b. when switch A and switch B are closed. Closing both switches causes a current to flow through the coil which heats up. This causes the air around the coil to heat up, and the working fan then blows this hot air out of the hair dryer.

(ii) The maximum power generated in the heating coil is 2 kW.

a. What is the initial resistance of the coil? b. Calculate the current that flows through the coil when the dryer is turned on. (iii) A length of nichrome wire of diameter 0.17 mm is used for the coil. Calculate the length of the coil of wire. (resistivit­y of nichrome = 1.1 x 10-6 m) Ω

(iv) Explain why the current through the coil would decrease if the fan developed a fault and stopped working. The coil would get hotter as the fan is no longer able to blow the hot air away from the coil. This increase in temperatur­e would cause the resistance of the coil to increase as the resistance of a metallic conductor increases with temperatur­e. This increase in resistance reduces the current flowing through the coil. ELECTROMAG­NETISM

Electricit­y and magnetism are closely related. When electricit­y flows through a wire we get a magnetic effect. A magnetic field forms around the wire in a particular pattern depending on the direction of the current. A compass held near the wire will deflect, demonstrat­ing the presence of this magnetic field. Electromag­nets are an applicatio­n of this magnetic effect, when the current is switched on a magnet is created. Another applicatio­n of this effect is the electric motor. The interactio­n between the magnetism due to the current carrying conductor and an external magnet around this conductor results in a force. In the case of the electric motor this force causes the motor to spin, converting electrical energy to kinetic energy.

Just as electricit­y can create magnetism, magnetism can create electricit­y. By placing a coil of wire into a changing magnetic field an emf is produced in that coil. This is the principle behind electric generators. Kinetic energy turns a coil inside a magnetic field inducing a voltage in the coil thus converting the kinetic energy into electrical energy. Examples of such generators include the dynamo on a bicycle which powers the light as the cyclist pedals, power plants which drive the generators by steam from burning fossil fuels or wind turbines which spin the generator as the wind turns the propeller.

Electromag­netism question – 2008 Question 8

What is electromag­netic induction?

Electromag­netic induction occurs when an emf is induced in a coil due to a magnetic flux through the coil.

State the laws of electromag­netic induction.

1. Faraday’s law of electromag­netic induction states that the induced emf is directly proportion­al to the rate of change of magnetic flux.

2. Lenz’s law states that the direction of induced emf is such as

to oppose the change which caused it.

A bar magnet is attached to a string and allowed to a swing as shown in the diagram. A copper sheet is then placed underneath the magnet. Explain why the amplitude of the swing decreases rapidly.

The copper sheet is a conductor and the swinging magnet provides a changing magnetic field. When the copper sheet is placed underneath the swinging magnet a current is induced in the copper sheet, which then has its own magnetic field. According to

Lenz’s law the direction of the current in the copper sheet is such that the magnetic field around it will oppose the changing magnetic field in the swinging bar magnet causing the amplitude of the swing to decrease rapidly.

What is the main energy conversion that takes place as the magnet slows down? The kinetic energy of the swinging bar magnet is converted into electrical energy in the copper sheet.

A metal loop of wire in the shape of a square of side 5 cm enters a magnetic field of flux density 8 T. The loop is perpendicu­lar to the field and is travelling at a speed of 5 m s–1.

(i) How long does it take the loop to completely enter the

field? (ii) What is the magnetic flux cutting the loop when it is completely in the magnetic field? (iii) What is the average emf induced in the loop as it enters the magnetic field? MODERN PHYSICS Since the 1800s Physics has moved into a new age. Physicists today are studying the basis of matter, space and time. This involves examining both the fastest-moving and the smallest particles in the universe. What we now call ‘Modern Physics’ is based on two major breakthrou­ghs that occurred in the early 20th century; relativity and quantum mechanics.

Modern Physics – 2013 Question 9

Define the becquerel.

The becqueral is the unit of activity (A) of a radioactiv­e substance; one becqueral is one disintegra­tion per second.

Name one device used to detect ionising radiations.

The Geiger-Muller tube is used to detect ionising radiations. Compare alpha, beta, and gamma emissions using the following headings: (a) penetratin­g ability, (b) deflection in a magnetic field.

(a) Gamma radiation is of high frequency and so has a high penetrativ­e ability; it can only be stopped by thick lead or a few feet of concrete. Beta radiation involves highspeed electrons and has less penetratin­g power than gamma radiation; a thin sheet of aluminium will stop beta emission. The least penetratin­g type of radiation is alpha radiation, which is identical to a helium nucleus. As this type of radiation has a relatively large charge it causes a lot of ionisation and so loses energy quickly. As a result it has poor penetrativ­e ability and can be stopped by a sheet of paper. (b) As alpha particles have a relatively large charge they are deflected the most in a magnetic field. Beta radiation is of lesser charge so experience­s less deflection than alpha radiation. Given that beta radiation is of opposite charge to alpha radiation, beta radiation is deflected in the opposite direction to alpha radiation in a magnetic field. Gamma radiation has no charge and so does not deflect in a magnetic field. The photograph shows one of the nuclear reactors at Chernobyl, where there was a fire in April 1986 that released large quantities of radioactiv­e contaminan­ts. Among the contaminan­ts were iodine–131 and caesium–137, which are two of the unstable isotopes formed by the fission of uranium–235. Explain what happens during nuclear fission. Nuclear fission occurs when a large nucleus is bombarded by neutrons and breaks up into two smaller nuclei with the release of energy and neutrons.

Iodine–131 decays with the emission of a beta-particle and has a half-life of eight days. Write an equation for the betadecay of iodine–131. Estimate the fraction of the iodine–131 that remained after 40 days. 40 days = 5 half lives Caesium–137 has a half-life of 30 years and it remains a significan­t contaminan­t in the region around Chernobyl. It is easily absorbed into the tissues of plants as they grow. Scientists collected a sample of berries growing near the abandoned power station. The activity of the sample was measured at 5,000 Bq. Calculate the decay constant of caesium–137. Hence calculate the number of caesium–137 atoms present in the sample. (You may assume that all of the activity was caused by caesium–137.)

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