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M.35/36

Pre-Leaving Certificate Examination, 2009

Physics

Marking Scheme

Ordinary Pg. 2 Higher Pg. 19

ªM.36MS=¬

2009 M.35/36_MS 2/40 Page 2 of 38

Dublin Examining Board

Pre-Leaving Certificate Examination, 2009

Physics

Ordinary Level Marking Scheme (400 marks)

SECTION A (120 marks) Answer three questions from this section. Each question carries 40 marks. 1. In an experiment to investigate the relationship between the fundamental frequency of a stretched string

and its length, the following data was collected:

l / m 0.90 0.71 0.65 0.59 0.56 0.47

f / Hz 256 325 353 389 412 486

(i) Draw a labelled diagram of the apparatus used in this experiment. Indicate on the diagram where the length was measured. (15)

– Labelled diagram (5 × 3m) Diagram must include: – sonometer – two bridges – stretched string – method of providing tension – length marked between 2 bridges ** Deduct 2 marks for each label omitted or incorrect. (ii) How was the frequency of the stretched string measured? (2 × 3m) (6)

– tuning fork / signal generator with magnet – resonance

(iii) Calculate length

1 for each data point. (13)

– 1.11, 1.41, 1.54, 1.69, 1.79, 2.13 (each 2 1/l correct) (3m)

Plot a graph of frequency against length

1 . Put length

1 on the horizontal axis (X-axis).

– axes labelled (3m) – 4 points plotted correctly (4m) – straight line (3m) (iv) What does the graph tell you about the relationship between frequency and length? (6)

Any 1: (6m) – frequency is proportional to 1/length / – frequency is inversely proportional to length

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2. A student carried out an experiment to measure the focal length of a convex lens. An illuminated object was placed at a distance u from the lens and the image distance v measured. This procedure was repeated for a number of object distances. The table shows the data recorded.

object distance, u / cm 20 25 30 35 40

image distance, v / cm 59 38 30 26 24

(i) Draw a labelled diagram of the apparatus used in this experiment. Indicate the distances u and v on the diagram. (12)

– Labelled diagram (4 × 3m) Apparatus must include: – convex lens – illuminated object – screen – correct labelling of u and v ** Deduct 2 marks for each label omitted or incorrect. (ii) State two observations the student could have made about the image formed when the object

was placed 20 cm from the lens, apart from the fact that the image is formed 59 cm from the lens. (2 × 3m) (6)

– inverted – image bigger than object / magnified (iii) Calculate an average value for focal length of the convex lens f using the above data and the formula

f1 =

u1 +

v1. (16)

– 0669.0591

2011

1=+=

f(3m) f1 = 1/0.0669 = 14.93 cm (3m)

0663.0381

2511

2=+=

f f2 = 1/0.0663 = 15.08 cm (3m)

0667.0301

3011

3=+=

f f3 = 1/0.0667 = 15.00 cm (3m)

0670.0261

3511

4=+=

f f4 = 1/0.0670 = 14.92 cm

0667.0241

4011

5=+=

f f5 = 1/0.0667 = 15.00 cm

– 5

54321 ffffffaverage++++= = 14.99 cm (4m)

** Accept (3m) for one correctly calculated 1/f and (3m) for one correctly calculated f. ** Accept (2 × 3m) for any two correctly calculated values of f. ** Do not penalise faverage if error in f1 , f2 , f3

f4 and f5 consistent.

** Accept final answer for faverage between 14.9 cm and 15.1 cm.

** Deduct 1 mark if units omitted or incorrect. ** Award marks for fraction form. (iv) Why did the student not place the object at 10 cm from the lens? (2 × 3m) (6)

– inside focal length, virtual image formed – can not capture on screen

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3. You carried out an experiment to verify Boyle’s law for a fixed mass of gas. (i) Draw a labelled diagram of the apparatus that could be used in this experiment. (12)

– Labelled diagram (4 × 3m) Apparatus must include: – enclosed volume of gas – method of measuring volume, e.g., scale – method of measuring pressure – method of varying pressure ** Deduct 2 marks for each label omitted or incorrect. (ii) What measurements were taken during the experiment? (2 × 3m) (6)

– measured the volume – (at a number of) different pressures (iii) How did you ensure that the mass remained fixed during the experiment? (4m) (4)

– gas kept enclosed (iv) What other quantity had to be kept constant during the experiment? Explain how this was

achieved. (3 × 3m) (9)

– temperature – after adjusting pressure – allow temperature to settle for a few minutes before taking volume reading (v) State Boyle’s law and explain how the data you collected could verify it. (3 × 3m) (9)

– pressure is inversely proportional – to the volume – PV is the same for each pair of readings / straight line of P plotted against 1/V

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4. A student completed an experiment to measure the resistivity of a nichrome wire. The following is an extract from his report.

I stretched the wire to remove kinks. I attached crocodile clips to the wire and measured the distance

between the crocodile clips. I inserted wires in the crocodile clips and connected the wires to an ________________ to measure the resistance of the wire. I measured the diameter of the wire at 5 different points along the wire using a _________________. I calculated the cross sectional area of the

wire using A=πr2. I calculated the resistivity of the wire using l

RA=ρ . The following is the table of

results that I obtained.

resistance of wire between crocodile clips / Ω 22.1

length of wire between crocodile clips / mm 790

diameter of the wire / mm 0.24 0.25 0.26 0.24 0.25 (i) Fill in the spaces in the student’s report. (2 × 3m) (6)

– ohmmeter – micrometer (ii) Why were the kinks removed from the wire? (4m) (4)

– more accurate measurement of the length (iii) Why did the student not measure the total length of the wire, only the distance between the

crocodile clips? (2 × 3m) (6)

– length measurement must be the same between two points as – resistance measurement (iv) Using the data in the table, calculate the cross-sectional area of the wire. (4 × 3m) (12)

– average diameter = 0.248 mm – average radius = 0.124 mm = 0.000124 m – area = =πr2= (3.14)(0.000124)2 – 4.83 × 10–8 m2 (v) Calculate the resistivity of nichrome. (6m + 3m) (9)

– 79.0

)1083.4)(1.22( 8−×==l

RAρ correct substitution

– = 1.35 × 10–6 (Ω m) (vi) Why was a metre stick not used to measure the diameter of the wire? (3m) (3)

– not accurate enough for small measurement

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SECTION B (280 marks) Answer five questions from this section. Each question carries 56 marks. 5. Answer any eight of the following parts (a), (b), (c), etc. ** Marks awarded for the eight best answers. (a) State Newton’s first law of motion. (4m + 3m) (7) – no force on object – no change in velocity / no acceleration (b) What is the momentum of a car of mass 200 kg travelling at a velocity of 70 m s–1? (P = mu)

(4m + 3m) (7)

– p = (200) (70) – 14000 kg m s–1 ** Accept full marks if correct answer shown without work. ** Deduct 1 mark if units omitted or incorrect. (c) What property of a musical note varies with (i) amplitude, (ii) frequency? (4m + 3m) (7)

– loudness – pitch (d) What is dispersion? (4m + 3m) (7)

– splitting up of light – into spectrum / constituent colours (e) Draw a diagram showing how an image is formed in a plane mirror. (4m + 3m) (7)

– two rays reflected obeying law of reflection – line of rays appearing to meet behind the mirror (f) What is the unit of capacitance? (7m) (7)

– farad (g) What is the total resistance of the circuit shown? (3m + 2m + 2m) (7)

– 31

61111

21+=+=

RRR

– 1/R = ½ – R = 2 Ω

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(h) Give two properties of the electron. (7) Any 2: (4m + 3m) – (small) mass / 9.1 × 10–31 kg / – negative charge 1.6 × 10–19 C / – orbits the nucleus outside the nucleus / – deflected by electric magnetic fields / etc. (i) What instrument is used to measure the density of a liquid? (7m) (7) – hydrometer (j) What is the function of a transformer? (4m + 3m) (7) – to increase or decrease the value – of a.c. voltage

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6. Define the moment of a force. (2 × 3m) (12)

– force multiplied by – distance (from fulcrum) What is a lever? (2 × 3m)

– rigid body that rotates – about a fixed point Explain why it is an advantage to have long handles on a wheelbarrow. (2 × 3m) (6)

– large moment of a force – due to large distance from the fulcrum / from end of handle Two children, one of mass 50 kg and the other of mass 30 kg, sit on either end of a seesaw, both 1.5 m

from the fulcrum.

(i) What is the weight of each child? (2 × 3m) (6)

– W = mg = (50)(9.8) = 490 N – W = mg = (30)(9.8) = 294 N (ii) What is the moment of each child about the fulcrum? (4 × 3m) (12)

– M = fd = (490)(1.5) – = 735 N m – M = fd = (294)(1.5) – = 441 N m ** Accept full marks if correct answer shown without work. ** Do not penalise error consistent with previous part. ** Deduct 1 mark if units omitted or incorrect. (iii) Where should the 50 kg child move to in order for the seesaw to be balanced? (9)

– 490x = 441 (3m + 3m) – x = 0.9 m (3m) (iv) If both children stay at the ends of the seesaw, where should a third child of mass 30 kg sit so

that the seesaw is balanced? (11)

– 735 = 294 x + 441 (3m + 3m) – 294x = 735 - 441 = 294 (3m) – x = 1 m (2m)

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7. When a person perspires, they are cooling down due to latent heat. (i) What is latent heat? (3 × 3m) (9)

– heat energy needed – to change state of substance – without a change in temperature (ii) Explain how perspiring helps to cool a person down. (2 × 3m) (6)

– water produced evaporates – takes latent heat from body (iii) What change of state occurs during perspiration? (3m) (3)

– liquid to gas / evaporation (iv) What is the difference between latent heat of vaporisation and latent heat of fusion? (9)

– heat energy required to change a substance from (3m) – fusion – a solid to a liquid (3m) – vaporisation – a liquid to a gas (3m)

(v) In the above diagram, which sections represent latent heat of vaporisation and latent heat

of fusion? (2 × 3m) (6)

– latent heat of fusion B – latent heat of vaporisation D A 200 g piece of ice at 0 °C is added to a container of water at 20 °C. The temperature of the water

eventually settled at 10 °C. Assuming that no heat is lost to the surroundings, (23) Calculate: (i) the heat required to melt the ice; (2 × 3m)

– Q = ml = (0.2)(3.3 × 105) – = 6.6 × 104 J (ii) the heat required to raise the temperature of the melted ice from 0 °C to 10 °C; (2 × 3m)

– Q = mc∆θ = (0.2)(4200)(10) – = 8400 J

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(iii) the heat lost by the water; (3m + 2m)

– same as heat gained by ice, 6.6 × 104 + 8400 – 74400 J (iv) the mass of the water. (2 × 3m)

– Q = mc∆θ, 74400 = m(4200)(10) – m = 1.77 kg ** Accept full marks if correct answer shown without work. ** Deduct 1 mark if units omitted or incorrect.

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8. What is meant by the refraction of light? State the laws of refraction of light. (15)

Refraction (2 × 3m) – bending of light – going from one medium to another

Laws of refraction (3 × 3m) – refracted ray, incident ray and normal – all in the same plane – sine of angle of incidence α sine angle of refraction / sin i ÷ sin r=constant A person looking at a still lake observes a fish under the water.

(i) Why does the fish appear to be less deep than it actually is? (3 × 3m) (9)

– light from fish bends at surface – away from normal – person thinks light is coming from that direction ** Accept diagram demonstrating above points. (ii) If the fish is at a depth of 2 m but it appears to be at a depth of 1.5 m, what is the refractive index

of water? (2 × 3m) (6)

– 5.1

2==dDn

– n = 1.33 A person scuba diving under water can only see through a certain circle out of the water due to

total internal reflection.

(i) What is total internal reflection? (3 × 3m) (9)

– light travels from more dense to less dense medium – angle greater than critical angle – all light is reflected back (ii) Describe an experiment to demonstrate total internal reflection. (3 × 3m) (9)

– diagram showing light beam and semicircular piece of glass – increase angle of incidence – beyond critical angle, all light is reflected in (iii) An optical fibre is an example of the application of total internal reflection. What is the smallest

angle that light must be incident on the inside surface of the optical fibre, if the light is to be totally internally reflected? The material of the optical fibre has a refractive index of 1.54. (8)

– SinC

n 1= , SinC

154.1 = / (3m)

– sin C = 0.649 / (3m) – C= 40.5° / (2m) ** Accept full marks if correct answer shown without work. ** Deduct 1 mark if units omitted or incorrect.

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9. What are magnetic field lines? (2 × 3m) (6)

– lines showing direction – north pole would travel in magnetic field Draw the magnetic field around a (12)

(i) Bar magnet (2 × 3m) – correct shape – correct direction of arrows (ii) Long straight wire carrying a current (2 × 3m) – correct shape – correct direction of arrows Describe an experiment that shows the magnetic field due to current in a long straight wire. (4 × 3m) (12)

– diagram showing wire passing through cardboard or something capable of supporting plotting compasses

– place plotting compass on cardboard and mark both ends of arrow – repeat until full circle is completed – repeat for a number of circles In an electric motor a magnetic field of flux density 4 T strikes the 10 cm side of the coil at right angles.

The coil carries a current of 5 A and is free to rotate around the axis shown. (i) What is the size of the force on either of the 10 cm sides of the coil? (9)

– F= BIl = (4)(5)(0.1) (correct substitution and conversion) (3m + 3m) – F = 2N (3m) (ii) Draw a diagram showing the direction of the forces on the coil. (2 × 3m) (6)

** Correct direction for each. (LHS = up, RHS = down). (iii) What is the size of the force is exerted on the 5 cm side of the coil? Explain your answer.

(2 × 3m) (6)

– no force – wire parallel to magnetic field (iv) In a practical motor, why is there a large number of turns in the coil? (2m + 3m) (5)

– size of force produced – multiplied by numbers of turns

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10. In a nuclear reactor, a large amount of energy is released by nuclear fission of uranium or plutonium. (12)

Explain the principle of nuclear fission. (3 × 3m)

– splitting up of a large nucleus – into two smaller nuclei – with a release of energy What particles are used to cause fission of uranium or plutonium?

Any 1: (3m) – of roughly the same size / by bombarding with neutrons / – neutrons cause splitting Explain the function of each of the following in a nuclear reactor: (15)

(i) Moderator (2 × 3m) – slows down neutrons – so they produce further fission

(ii) Control rods (2 × 3m) – block the neutrons – can speed up or slow down the reaction

(iii) Shielding (3m) – stops radiation escaping / protect operator from radiation State two ways that nuclear fusion has an adverse effect on the environment. (6)

Any 2: (2 × 3m) – mining of uranium releases radon / – very long half-life of radioactive waste / – transport of spent fuel rods to reprocessing / – escape of radioactivity from reactor / etc. The sun’s principal source of energy is nuclear fusion. The amount of heat produced is sufficient to

keep the reaction going and the amount of fuel available will last for billions of years. What is nuclear fusion? (3 × 3m) (9)

– joining or fusing – two small nuclei – to form large nucleus / release of energy What gas is the primary fuel for fusion in the sun? What gas is produced as a result of fusion? (6)

Primary fuel (3m) – hydrogen

Gas produced as a result of fusion (3m) – helium How does E = mc2 explain where the energy released from the sun comes from? (3m + 3m + 2m) (8)

– loss in mass m – when 2 hydrogen atoms combine to produce 1 helium atom – converted to large amount of energy

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11. Read this passage and answer the questions below. At any given moment, there are 1,800 thunderstorms in progress somewhere on the earth. A

thunderstorm forms in air that has three components: moisture, instability and something such as a cold front to cause the air to rise. Continued rising motions within the storm may build the cloud to a height of 35,000 to 60,000 feet (6 to 10 miles) above sea level. Temperatures higher in the atmosphere are colder; ice forms in the higher parts of the cloud. Ice in a cloud seems to be a key element in the development of lightning. Storms that fail to produce quantities of ice may also fail to produce lightning.

In a storm, the ice particles vary in size from small ice crystals to larger hailstones, but in the

rising and sinking motions within the storm there are a lot of collisions between the particles. This causes a separation of electrical charges. Positively charged ice crystals rise to the top of the thunderstorm, and negatively charged ice particles and hailstones drop to the middle and lower parts of the storm. Enormous charge differences (electrical differential) develop.

(Adapted from the US National Weather website)

(a) Fahrenheit is one unit of temperature. Name two other units of temperature. (7)

Any 2: (4m + 3m) – degrees Celsius – Kelvin (b) Why do positive and negative charges position themselves at opposite sides of the cloud? (3m + 4m) (7)

– opposite charges – repel each other (c) Draw a diagram showing the electric field produced due to a positive and a negative charge near

each other. (4m + 3m) (7)

– correct shape – arrows from positive to negative (d) Charge is placed on the conducting solid below. Show how the charge is distributed on this conductor. (4m + 3m) (7) – charge on outside of conductor – more concentrated at pointed end (e) Why should you stop playing golf during a lightning storm? (4m + 3m) (7)

– metal pointed surface – acts as conducting path for charge in clouds (f) Why are lightning conductors placed on buildings? (4m + 3m) (7)

– lightning will be conducted through pointed conductors – rather than through buildings

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(g) Name two quantities on which the force between two charges depends. (7)

Any 2: (4m + 3m) – size of the charges / – distance between charges / – permittivity of medium (h) Why should a person working with integrated circuits be connected to Earth? (4m + 3m) (7)

– static build up on person’s body – could discharge through and damage circuits

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12. Answer any two of the following parts (a), (b), (c), (d). (a) Define energy. (2 × 3m) (6)

– ability – to do work What energy conversion takes place in an electric motor? (2 × 3m) (6)

– electric energy – to kinetic energy A cannon ball of mass 20 kg is fired vertically upwards from a cannon so that the ball rises to a

height of 50 m. What is the potential energy gained by the ball? (2 × 3m) (6)

– P.E.= mgh = (20)(9.8)(50) – = 9800 J At what velocity did the ball leave the cannon? (3m + 2m + 2m) (7)

– v2 = u2 + 2as, 02 = u2 + 2(–9.8)(50) – u2 = 980 – u = 31.3 m s–1 At what speed does the ball hit the cannon on the way down? (3m) (3) – speed = 31.3 m s–1 (b) Define (i) activity and (ii) half-life of a radioactive substance. (12)

Activity (2 × 3m) – number of nuclei decaying – per second

Half-life (2 × 3m) – time for half (of undecayed atoms) – to decay State one use of radioisotopes. (4)

Any 1: (4m) – any medical application / – carbon dating / – smoke detectors / – tracers / – sterilising food / etc. The half-life of a radioactive isotope is 2 years. (i) How many half lives are there in 10 years? (3m) (3)

– 5

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(ii) What fraction of the original sample remains after 10 years? (3m) (3) – 1/32 (iii) How long does it take so that the sample has decayed to 1/8 of its original size? (2 × 3m) (6)

– ½ 2 years, ¼ 4 years – 1/8 6 years ** Full 6m marks if correct answer written directly. (c) In Ireland, mains electricity is supplied to homes at 230 V a.c.

What is an electric current? (2 × 3m) (6)

– flow of / movement – charge / electrons /electricity An electric fire has a power rating of 3 kW when connected to mains voltage. The fuse in the plug

was found to be a 5 A fuse. Calculate the current that flows when the electric fire is first plugged in. (9)

– P=VI or 3000=230(I) (3m)

– I=230

3000 (3m)

– I=13 A (3m) With high voltage present in a house circuit several safety precautions must be taken. Miniature

circuit breakers and residual circuit breakers are devices often found in a house. Explain the principle on how one of these circuit breakers work. (7)

MCB (4m + 3m) – bimetallic strip and electromagnet – large current causes switch to trip or RCB (3m + 2m + 2m) – detects difference between current in live and in neutral – when someone makes contact with live – trips switch If mains voltage is supplied to a heater of resistance 50 Ω, what is the a.c. current flowing through

the heater? (2 × 3m) (6)

– I = V/R= 230/50 – 4.6 A (d) Light waves are transverse, electromagnetic waves while sound waves are longitudinal, mechanical waves.

(i) What is the difference between transverse and longitudinal waves? (3 × 3m) (9)

– transverse vibrate perpendicular – longitudinal vibrate parallel – to direction of motion

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(ii) Name two other electromagnetic waves. (6)

Any 2 (2 × 3m) – radiowaves / – microwaves / – infrared / IR waves / – ultraviolet /UV waves / – X-rays / – γ-rays / etc. (iii) What is a polarised wave? (2 × 3m) (6)

– vibrates – in only one plane (iv) Sound waves can not be polarised. Explain why. (4m) (4)

– vibrates along a line / longitudinal (v) Give an application of polarisation. (3)

Any 1: (3m) – sunglasses / – cameras / – stress measurement / etc.

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Dublin Examining Board

Pre-Leaving Certificate Examination, 2009

Physics

Higher Level Marking Scheme (400 marks)

SECTION A (120 marks) Answer three questions from this section. Each question carries 40 marks. 1. In an experiment to investigate the relationship between the force F applied to a trolley and its

acceleration a, the following results were obtained.

F / N 0.5 1.0 1.5 2.0 2.5 3.0

a / m s–2 0.82 1.63 2.46 3.26 4.07 4.89 Describe how the force was applied and the acceleration was measured. (12) Force applied using (3m) – hanging masses attached to trolley / other method of accelerating object, e.g., newtonmeter

Acceleration measured using Any 3: (3 × 3m) – ticker timer and ticker tape / light gate and timer / – measurement of initial and final velocity and time between both measurements / – measurement of initial and final velocity and distance between both measurements / – calculation of acceleration using v = u +at / v2=u2+2as Draw a suitable graph showing the relationship between the applied force and acceleration. Use your graph to calculate the mass of the accelerating objects. (18)

Graph – Axes labelled correctly (3m) – At least 6 points correctly plotted (6 × 1m) – good distribution (about straight line) (3m) ** (–1m for each incorrect point plotted)

Mass of accelerating objects (2 × 3m) – measure slope of graph (taking 2 points from line, not data) – correct value = 0.61 kg (0.60 to 0.62) Why was it important that the mass of the system was kept constant? Explain how this was achieved. (10) Mass kept constant (4m) – force is only proportional to acceleration if mass is constant / only two variables

How achieved (2 × 3m) – as mass of system includes hanging masses – transfer masses from scale pan to trolley (to keep mass of system constant)

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2. In an experiment to verify Snell’s law, a student measured the angle of incidence i and the corresponding angle of refraction r for light entering a glass block. This was repeated for a number of different angles of incidence. The following data was recorded.

i / degrees 15 25 35 45 55 65

r / degrees 9 17 23 29 34 38 Describe, with the aid of a diagram, how the student measured the angle of refraction. (9)

– Labelled diagram (3 × 3m) Diagram must include: – pins / raybox (to obtain incident and refracted rays) – outline of block, incident and refracted ray, normal – measure angle between refracted ray and normal using a protractor ** Deduct 2 marks for each label omitted or incorrect. Draw a suitable graph and explain how it verifies Snell’s law. (15)

Graph (4 × 3m) – calculate sin i and sin r

sin i 0.26 0.42 0.57 0.71 0.82 0.91

sin r 0.16 0.29 0.39 0.48 0.56 0.62

– axes labelled – at least 5 points correctly plotted – straight line drawn

Verifies (3m) – Snell’s law states sin i is proportional to sin r / straight line through origin Use your graph to calculate the critical angle of the glass block. (3 × 3m) (9)

– measure slope of graph (taking two points from line, not from data) e.g., m=y2–y1/x2–x1 – slope = n = 1.46 (1.44 to 1.48) – n = 1/sin C, C = 43° Which result is the least accurate? Explain your answer. (3m + 4m) (7)

– angle of incidence = 15° / angle of refraction = 9° – small angle of incidence and refraction, higher percentage error, more difficult to

measure accurately

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3. In an experiment to investigate the relationship between the fundamental frequency f of stretched string and its tension T, a student recorded the following data. The length of the string was kept constant at 80 cm.

T / N 10 16 19 23 26 36

f / Hz 256 325 353 389 412 486 Describe how the tension of the string was varied and measured. (2 × 3m) (6)

– tension changed using hanging mass / newtonmeter or spring balance attached to wire – tension is weight of hanging mass / reading on spring balance / newtonmeter How did the student know that the string was vibrating at its fundamental frequency and not at a

different harmonic? (3 × 3m) (9)

– set length of wire near maximum and set at minimum tension – increase tension, until resonance with lowest frequency tuning fork – paper rider jumps from wire when resonance Draw a suitable graph and explain the relationship between the frequency and the tension

of the string. (15)

Graph – calculate square root of tension, values of T T=3.16, 4, 4.36, 4.80, 5.10, 6 (3m) – axes labelled (3m) – 6 points correctly plotted (3m)

Relationship (2 × 3m) – straight line through the origin – frequency proportional to square root of tension From your graph, find the fundamental frequency when the tension was 20 N and the length was

reduced to 40 cm. (3m + 3m + 4m) (10)

– 47.420 = – from graph f = 362 Hz – when length changes to 40 cm, f = 724 Hz

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4. A student investigated the relationship between current I and potential difference V for a copper sulfate solution.

V / V 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

I / mA 56 87 114 145 161 201 231 248 291 Explain, using a labelled diagram, how the data was collected. (12)

– Labelled diagram (4 × 3m) Diagram must include: – method of varying voltage – ammeter in series, voltmeter in parallel across electrodes – electrodes in solution – vary voltage and take current reading and voltage reading ** Deduct 2 marks for each label omitted or incorrect. Draw a suitable graph to demonstrate that a copper sulfate solution obeys Ohm’s law. (15)

Graph – axes labelled (3m) – 6 points correctly plotted (3m) – straight line, good fit (3m)

Obeys Ohm’s law (2 × 3m) – Ohm’s law, current proportional to voltage – verified by straight line through the origin What are suitable electrodes for this experiment? If inactive electrodes (e.g. graphite) were used instead, how would the shape of the graph differ? Explain why the two graphs differ in shape. (13)

Electrode (3m) – copper

Graph shape different (either stated or drawn) (2 × 3m) – straight line graph – V > 0, when I = 0 / current increases at V > 0

Why graphs differ in shape (2 × 2m) – for inactive electrode – behaves as a cell / produces EMF – opposes applied voltage / applied voltage must overcome EMF for current to flow

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SECTION B (280 marks) Answer five questions from this section. Each question carries 56 marks. 5. Answer any eight of the following parts (a), (b), (c), etc. ** Marks awarded for the eight best answers. (a) State Boyle’s law. (4m + 3m) (7)

– pressure inversely proportional to volume – at constant temperature for a fixed mass of gas (b) A ship that can travel at 50 m s-1 in still water aims in a northerly direction. If it meets a current

of 15 m s-1 going east, what is the resultant speed and direction of the boat? (4m + 3m) (7)

– speed = 22 1550 + = 52.2 m s–1 – tan θ = 15/50, θ = 16.7° ** Deduct 1 mark if units omitted or incorrect. ** Accept full marks if correct answer shown without work. (c) Define sound intensity. (4m + 3m) (7)

– rate at which sound energy is passing – through unit area at right angles (d) What is the principle on which residual current devices are based? (4m + 3m) (7) – detects difference between current in live and in the neutral – when above pre-set value, RCD trips / disconnects circuit from live (e) What speed would light travel in an optical fibre whose refractive index is 1.54? (3m + 4m) (7)

– n= c/v – v = c/n = (3 × 108)/1.54 = 1.95 × 108 m s–1 ** Deduct 1 mark if units omitted or incorrect. ** Accept full marks if correct answer shown without work. (f) Calculate the angular velocity of an object on the surface of the earth, due to the rotation of

the earth about its axis. (3m + 2m + 2m) (7)

– ω= 2π/T – = 2π/(24×60×60) – 7.27 × 10–5 rad s–1 ** Deduct 1 mark if units omitted or incorrect. ** Accept full marks if correct answer shown without work. (g) Why would a mercury thermometer and a resistance thermometer give different values for the

temperature of a room? (4m + 3m) (7)

– different thermometric properties – vary differently with temperature

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(h) Why is the capacitance of a parallel plate capacitor greater than that of a single charged plate? (3m + 4m) (7)

– oppositely charged plate reduces potential – as C = Q/V, capacitance increases (i) What is the maximum kinetic energy of an electron in a cathode ray tube when the voltage

between the cathode and the anode is 2000 V? (4m + 3m) (7)

– kinetic energy gained = eV – = (1.6 × 10–19)(2000) = 3.2 × 10–16 J ** Deduct 1 mark if units omitted or incorrect. ** Accept full marks if correct answer shown without work. (j) Write down the quark structure of an anti-proton. (7m) (7)

– duu or State the principle on which the moving coil loudspeaker is based. (4m + 3m) (7)

– current carrying conductor in a magnetic field – experiences a force

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6. State Newton’s second and third laws of motion. (12)

Newton’s second law (2 × 3m) – net force is proportional to / F = ma – rate of change of momentum / explain notation

Newton’s third law (2 × 3m) – every action / force exerted by A on B is equal – has an equal and opposite reaction / and opposite to force exerted by B on A Explain how Newton’s first law is a special case of Newton’s second law. (2 × 3m) (6)

– when applied force = 0 – no change in momentum / velocity remains constant Using Newton’s third law, explain why a person in a lift, moving vertically upwards, exerts a greater

force on the lift floor when the lift starts moving upwards from rest than when the lift is slowing to rest. (3 × 3m) (9)

– when accelerating upwards, net force is upwards / there is an extra force caused by the lift’s acceleration

– normal reaction greater than weight / person appears temporarily heavier – when decelerating, net force downwards, weight greater than normal reaction / person appears

lighter as the lift is tending to fall away from the person A crane cable that is capable of withstanding 22,000 N is attached by a hook to a 2,000 kg block that is

resting on the ground. The cable initially starts lifting the block at the maximum acceleration that the cable can withstand for 2 seconds. It then continues to raise the block at constant velocity for a further 5 seconds. At this time the block slips off the hook at the end of the cable. (29)

Calculate: (i) the tension in the cable when the block is moving at constant velocity; (2 × 3m)

– T = mg – T = (2000) (9.8) = 19600 N ** Deduct 1 mark if units omitted or incorrect. (ii) the maximum acceleration that the cable can withstand; (6m + 3m)

– T – mg = ma – 22000–19600 =2000a, a = 1.2 m s–1 ** Deduct 1 mark if units omitted or incorrect. (iii) the maximum height that the block reaches above the ground.

– when accelerating v = u + at, 0 +(1.2)(2) = 2.4 m s–1 (3m) – S = ut + ½ at2 = 0(2) + ½ (1.2)(2)2 = 2.4 m (3m) – constant velocity s = ut = (2.4)(5) = 12m (2m) – after slipping off, v2 = u2 + 2as, (0)2 = (2.4)2 +2(-9.8)s, s = 0.29 m (3m) – total distance = 12 + 2.4 + 0.29 = 14.69 m (3m) ** Deduct 1 mark if units omitted or incorrect.

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7. Constructive interference occurs when two coherent sources are used. Explain what is meant by constructive interference and coherent sources. (12)

Constructive interference (2 × 3m) – when two waves meet, amplitude of resultant wave / crest (trough) of wave meets crest

(trough) of another wave – of greater amplitude than any individual wave amplitudes

Coherent sources (2 × 3m) – same frequency / wavelength – in phase or constant phase difference Describe an experiment to demonstrate constructive interference for sound waves. (4 × 3m) (12)

– 2 loudspeakers connected to signal generator / vibrating tuning fork rotated – walk in line parallel to line between speakers / close to ear – hear sounds louder than sounds from individual loudspeaker / tuning fork at some points – constructive interference occurs State two uses of microwaves. (6)

Any 2: (2 × 3m) – ovens / – remote controls / – television signals / – any other relevant application Two single frequency coherent microwave beams are directed from the same point towards a

concave surface that reflects microwaves and is of focal length 20 cm. One beam is incident parallel to the principal axis of the concave reflector, 6 cm from the axis. The other beam passes through the focal point before striking the mirror. The strongest possible signal that could be produced by the combination of the two beams is detected where they meet, 30 cm from the concave surface. (26)

Calculate: (i) how far from the concave surface the two beams originated; (2 × 3m)

– fvu111 =+

201

3011 =+

u

– u = 60 cm (ii) how far below the axis the two beams met; (2 × 3m)

– m = v/u = 30/60 = 0.5 – distance below axis = 6(0.5) = 3 cm

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(iii) the minimum possible frequency of the microwaves.

– distance travelled by beam incident parallel = 22 93060 ++ = 91.32 cm (4m) – distance travelled by beam incident through focus =

22 96030 ++ = 90.67 cm (4m) – path difference = 0.65 = 1λ / (4m) – f = c/ λ = (3 × 108)/(0.0065) = 4.62 × 1010 Hz (2m)

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8. Define (i) potential difference, (ii) electromotive force. (12)

Potential difference (2 × 3m) – work done / electrical energy converted to other forms / V=W/Q – moving unit charge between two points / which 1 C passes between 2 points / notation

Electromotive force (2 × 3m) – voltage applied / amount of non-electrical energy converted into electrical energy – to a circuit / when a charge of 1 C flows around a circuit Two resistors R1 and R2 are connected in parallel. These in turn are connected in series with a third

resistor R3.

Derive an expression for the resistance of a single resistor that could replace the two resistors in parallel and still produce the same current flowing through R3, when a voltage supply is connected across the circuit.

Derive an expression for the effective resistance of the entire circuit. (18)

Parallel part of the circuit (4 × 3m) – ITotal = I1 + I2

– 21 R

VRV

RV

P+=

– 21

111RRRP

+=

– 21

21

RRRRRP +

=

** Deduct 1 mark if units omitted or incorrect.

Parallel part in series with R3 (2 × 3m) – VTotal = IRp + IR3

– RTotal = Rp +R3 = 321

21 RRR

RR ++

** Deduct 1 mark if units omitted or incorrect. Draw a circuit diagram of a Wheatstone bridge, with resistors 10 Ω and 15 Ω along the top branch of the

circuit and resistors 20 Ω and 35 Ω along the bottom branch. (6)

– Diagram (2 × 3m) Diagram must include: – correct alignment of 4 resistors – galvanometer correctly located (i) If 12 V is connected across the Wheatstone bridge, calculate the current flowing through each resistor. (6)

Top branch (3m) – I1 = 12/(10 + 15) = 0.48 A

Bottom branch (3m) – I2 = 12/(20 + 35) = 0.22 A

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(ii) If a galvanometer is connected between the middle of the top branch and the middle of the bottom branch of the circuit, in what direction will the current flow? Explain your answer. (11)

– potential difference across 10 Ω resistor = 0.48(10) = 4.8 V (3m) – potential at top of galvanometer = 12 - 4.8 = 7.2 V / (2m) – potential difference across 20 Ω resistor = 22(20) = 4.4 V – potential at bottom of galvanometer = 12 - 4.4 = 7.6 V (3m) – current flows from bottom to top / from high to low potential (3m) ** Accept different answer, for correct method employed, if resistors in diagram were

arranged in a different order. (iii) How can the Wheatstone bridge be balanced by replacing only one resistor with another? (3m) (3)

– replace 35 Ω with 30 Ω resistor or any other correct switch

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9. State three factors that affect the force exerted on a current carrying conductor in a magnetic field. (3 × 3m) (9)

– magnetic flux density – length of conductor – current flowing through conductor Sketch the shape of the magnetic field due to the current in a long straight wire. (6)

Sketch (2 × 3m) – concentric circles around wire – correct direction of field lines according to right hand grip rule Define the ampere. (3 × 3m) (9)

– two (infinitely) long parallel wires (of negligible cross sectional area) – 1 m apart in a vacuum – experience a force of 2 × 10–7 N per unit length A rectangular coil of wire is free to rotate about an axis through its centre as shown. It is placed in a

uniform magnetic field of flux density 12 T. The direction of the magnetic field is in the plane of the coil and is perpendicular to two sides of the coil. The force exerted on the 6 cm length side of the coil is 2 N. (23)

Calculate: (i) the current flowing in the coil;

– F=BIl / 2=12I(0.06) (4m) – I = 2.78 A (4m) ** Deduct 1 mark if units omitted or incorrect. (ii) the moment of the force exerted on the coil about the

axis; (2 × 3m)

– moment of couple = F × perpendicular d – 2 × 0.02 = 0.04 N m ** Deduct 1 mark if units omitted or incorrect. (iii) the moment of the force when the coil has rotated through 45°; (2 × 3m)

– perpendicular distance = 0.02 cos 45 = 0.014 N m – moment = 2 × 0.014 = 0.028 N m ** Deduct 1 mark if units omitted or incorrect. (iv) how the moment of the force change as the coil rotates through 90°. (3m)

– as it rotates through 90°, the moment changes direction / becomes zero as cos 90° = 0 The coil rotates through 90° in 0.1 seconds. What is the average linear velocity of a point on either of the 6 cm length sides of the coil? (3 × 3m) (9)

– average angular velocity (ω) = π/2/0.1 = 15.57 rad s–1 – v = ωr – v = (15.57)(0.02)=0.311 ** Deduct 1 mark if units omitted or incorrect.

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10. Answer either part (a) or part (b).

(a) An electron is emitted from a nucleus during beta decay. Write down the nuclear equation for this decay. (3m + 3m + 3m + 1m) (10)

– ν00

01

11

10 ++→ − epn

** Maximum 4 marks if no numbers. ** Deduct 1 mark if neutrino is omitted. Name the family of particles to which electrons belong. (3m) (3)

– leptons Read the following passage and answer the accompanying questions.

In 1928 Paul Dirac proposed the existence of a particle that had the same mass as an electron but with a positive charge. In 1932, Carl David Anderson of the California Institute of Technology noticed a peculiar track in a cloud chamber. A cloud chamber shows the path of charged particles that pass through the chamber deflected by magnetic fields. In the chamber, there was a circular track that looked just like the track of an electron, but it curved with the same radius in the opposite direction. Anderson concluded it was a particle with the mass of an electron and the charge of a proton.

Adapted from “Probing the Proton”; Parker; 2008 (i) What name was given to the “positively charged electron” described? (3m) (3)

– positron (ii) What name is given to the process that produced two particles in the cloud chamber? (3m) (3)

– pair production (iii) Why did the two particles travel in a circular path and in opposite directions? (2 × 3m) (6)

– force on a charged particle at right angles to direction of travel – oppositely charged particles deflected in opposite direction / left hand rule (iv) Why could the positively charge particle observed in the chamber not be a proton? (3m) (3)

– particles of different mass would travel in circles of different radii (v) Write a reaction to represent the production of the two particles in the chamber. (6m) (6)

– hf mc2 + mc2 / – γ e- + e+ (vi) Calculate the maximum wavelength of the γ-ray photon required to produce the

particles. (4 × 3m) (12)

– hf = 2 mc2 – (6.6 × 10-34)f = 2(9.1 × 10–31)(3.0 × 108)2 – 2.5 × 1020 Hz – λ = c/ f = 1.2 × 10–12 m ** Deduct 1 mark if units omitted or incorrect.

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(vii) If a γ-ray, with the minimum frequency needed to produce an electron and its antiparticle, is incident on the cloud chamber, why are no tracks observed? (2 × 3m) (6)

– all energy of photon used in producing mass of particles – no energy left over to give kinetic energy to particles (viii) A positively charged particle is not produced on its own. Explain why. (4m) (4)

– conservation of charge (b) Describe how n-type and p-type silicon semiconductors are produced. Explain the significance

of the terms n-type and p-type. (12)

Produced (2 × 3m) – p-type, silicon doped with boron – n-type, silicon doped with phosphorus

Significance (2 × 3m) – n-type, excess of negatively charged free electrons over positive holes – p-type, excess of positive holes over free electrons Using a diagram, explain the operation of a p-n junction in both forward biased and reverse

biased. (4 × 3m) (12)

– formation of depletion layer – production of junction voltage – reverse biased, widening of depletion layer, no current flows – forward biased, narrowing of depletion layer, current flows Draw circuit diagrams showing how a.c. voltage can be converted to d.c. voltage using (i) a half

wave rectifier, (ii) a bridge rectifier. (14)

Half wave rectifier (2m + 3m) – a.c. supply, d.c. output – in series with semiconductor diode

Bridge rectifier (3 × 3m) – a.c. supply connected correctly to – diamond shape of 4 diodes – correct orientation of diodes Sketch the output voltage produced by (i) the half wave rectifier, (ii) the bridge rectifier. (12)

Half wave rectifier (2 × 3m) – truncated sine wave – above or below the line only

Bridge rectifier (2 × 3m) – truncated sine wave – no gaps in wave pattern (accept either smoothed or unsmoothed wave) What happens in a photodiode when light shines on it? (2 × 3m) (6)

– reverse biased p-n junction – light creates electron-hole pairs / current flows when light shines on it, current flowing

is directly proportional to the intensity of light

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11. Read the following passage and answer the accompanying questions. Two mysterious discoveries led Marie Curie to her life’s work. In December 1895, a German

physicist, Wilhelm Roentgen, had discovered rays that could travel through solid wood or flesh. A few months later a French physicist, Henri Becquerel, discovered that minerals containing uranium also gave off rays. Roentgen’s X-rays amazed scientists, who took to studying them with great energy. They mostly ignored Becquerel’s rays, which seemed much the same, only weaker. Marie decided to investigate the uranium rays.

She started off by studying a variety of chemical compounds that contained uranium and discovered that the strength of the rays that came out depended only on the amount of uranium in the compound. It had nothing to do with whether the material was solid or powdered, dry or wet, pure or combined with other chemical elements. If you had a certain amount of uranium, i.e., a certain number of uranium atoms, then you got a certain intensity of radiation. Nothing else made a difference.

(Adapted from “Marie Curie – Her Story in Brief”; The American Institute of Physics; 2008) (a) What is the nature of Roentgen’s X-rays? (3m + 4m) (7)

– high energy – electromagnetic radiation (b) How do α-rays and γ-rays differ from X-rays? (4m + 3m) (7)

– α-rays, particles / He nuclei – γ-rays, higher energy than X-rays

(c) 23892 U undergoes three alpha decays and two beta decays. What is the atomic number and mass

number of the element produced? (4m + 3m) (7)

– mass number = 226 – atomic number = 88 (d) “Marie Curie…discovered that the strength of the rays that came out depended only on the

amount of uranium in the compound”. How is this statement related to the law of radioactive decay? (4m + 3m) (7)

– activity / number of decays per second is proportional / NdtdN ∝

– to number of undecayed particles (e) Define the unit named after Henri Becquerel and state the quantity for which it is a unit.

(4m + 3m) (7)

– becquerel – one disintegration per second – rate of decay / activity

(f) 23592 U emits α-particles and has a half life of 7 × 108 years. How many α-particles are emitted

per second from a sample of U-235 that contains 2 × 1020 atoms? (4m + 3m) (7)

– λ = ln2/T1/2= 0.693/(7 × 108)(365)(24)(60)(60) = 2.2 × 1016 = 3.15 × 10–17 s–1 – rate of decay = (3.15 × 10–17)( 2 × 10 20) = 6300 decays per second ** Deduct 1 mark if units omitted or incorrect.

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(g) Compare the ionising and penetration properties of α-rays and γ-rays. (4m + 3m) (7)

– α-rays more ionising than γ-rays – γ-rays more penetrating than α-rays (h) Marie Curie suffered health problems, eventually leading to her death, as a result of long term

exposure to the radioactive element radium. State two ways in which radiation is a health hazard. (7)

Any 2: (4m + 3m) – any form of cancer / – genetic defects passed on / – death / – skin burns / etc.

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12. Answer any two of the following parts (a), (b), (c), (d). (a) Define simple harmonic motion. (3 × 3m) (9)

– acceleration of object is proportional – to displacement from equilibrium / fixed point – and is directed towards equilibrium / fixed point An object of mass 2 kg is executing simple harmonic motion. It experiences a force of 8 N

when the potential energy of the object is a maximum. The next time the object has maximum potential energy is 1 second later. (19)

Calculate: (i) the period of oscillation; (3m)

– 1 × 2 = 2 s (ii) the amplitude of the motion; (3m + 2m + 2m + 3m)

– 14.32

22 === ππωT

– a = F/m = 8/2 = 4 N – a=ω2A – 4 = (3.14)2A, A = 0.41 m (iii) the force felt by the object when it is midway between the positions of maximum and

minimum potential energy. (2 × 3m)

– a = ω2A = (3.14)2(0.205) = 2.38 m s–2 – F = ma = (2)(2.38) = 4.76 N (b) Define heat capacity. (2 × 3m) (6)

– energy needed to – change the temperature by 1 K /1 °C State the principle on which a storage heater is based. (2 × 3m) (6)

– contain bricks of high specific heat capacity – heated at night, gives out heat slowly during the day Calculate the heat capacity of a cylindrical piece of copper of height 50 cm and radius

30 cm. (3m + 3m + 2m) (8)

– volume = πr2h = (3.14)(0.30)2(0.5) = 0.1413 m3 – mass = density × volume = (8920)(0.1413) = 1260.396 kg – C = cm = (390)(1260.396) = 491554.44 J K–1 ** Deduct 1 mark if units omitted or incorrect.

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The piece of copper at 25 °C is placed in 100 kg of water at 80 °C in an insulated bath. Calculate the final temperature of the water. (3m + 3m +2m) (8)

– C(T – 25) = mc (80-T) – (491554.44)(T-25) = (100)(4200)(80 –T) / 491554.44T–12288861=33600000–420000T / 911554.44T=45888861 – T = 50.34132794°C ≅ 50.34°C ** Deduct 1 mark if units omitted or incorrect. (c) State Lenz’s law. (2 × 3m) (6)

– direction of induced current – opposes changes which produced it Explain why violating Lenz’s law violates the law of conservation of energy. (2 × 3m) (6)

– induced current flows in the opposite direction, causes a greater induced current and an increased induced emf

– work is done on the magnet and electrical energy is expended in the circuit, i.e. energy coming from nowhere – violates law of conservation of energy

A magnet is moved at 8 m s–1 towards a coil of wire of resistance 10 Ω. If the magnet experiences a force of 0.1 N, calculate the current induced in the coil due to the

moving magnet. (3m + 3m + 4m) (16)

– work done in moving magnet in 1 second = F × V = 0.1 × 8 = 0.8 J per sec – electrical energy dissipated in 1 second = I2R = I2(10) – 10I2 = 0.8, I = 0.28 A Why can the current flowing in the coil be different to this calculated value? (2 × 3m)

– changing current in coil produces changing magnetic field – induces opposing current / self induction (d) What is the photoelectric effect? (2 × 3m) (6)

– emission of electrons from a metal – by electromagnetic radiation of a suitable frequency Describe an experiment to demonstrate the photoelectric effect. (3 × 3m) (9)

– positively / negatively charged electroscope with zinc plate on cap – UV light shone on zinc leaves diverge / leaves collapse – when visible light shone on zinc, no effect

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Photons of wavelength 200 nm are incident on a zinc plate whose work function is 4.3 eV. (13) What is the energy of the photons? (3m)

– E = hc/λ = (6.6 × 10–34)(3 × 108)/200 × 10–9) = 9.9 × 10-19 J What is the maximum kinetic energy of the emitted photons? (2 × 3m)

– work function (Φ) = 4.3 × 1.6 × 10–19 = 6.88 × 10–19 J – max K.E. = hf – Φ = (9.9 ×10-19) – (6.88 × 10–19) = 3.02 × 10–19 J Why are all electrons not emitted at this maximum kinetic energy? (4m)

– electrons have different bonding energy / most loosely bound have highest kinetic energy

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