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L.35/36

Pre-Leaving Certificate Examination, 2018

Physics

Marking Scheme

Ordinary Pg. 4

Higher Pg. 29

2018.1 L.35/36_MS 2/60 of 59 examsDEB


Physics

Ordinary & Higher Levels

Table of Contents

Ordinary Level Higher Level Section A Section A

Q.1 .................................................. 4 Q.1 .................................................. 29

Q.2 .................................................. 6 Q.2 .................................................. 31

Q.3 .................................................. 7 Q.3 .................................................. 32

Q.4 .................................................. 8 Q.4 .................................................. 34

Section B Section B Q.5 .................................................. 10 Q.5 .................................................. 36

Q.6 .................................................. 13 Q.6 .................................................. 39

Q.7 .................................................. 15 Q.7 .................................................. 41

Q.8 .................................................. 17 Q.8 .................................................. 43

Q.9 .................................................. 19 Q.9 .................................................. 46

Q.10 .................................................. 21 Q.10 .................................................. 48

Q.11 .................................................. 23 Q.11 .................................................. 50

Q.12 .................................................. 25 Q.12 .................................................. 52

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2018.1 L.35/36_MS 3/60 of 59 examsDEB


Physics

Ordinary & Higher Levels

Explanation

Conventions Used

1. A dash – before an answer indicates that the answer is a separate answer, which may be considered as independent of any other suggested answers to the question.

2. A forward slash / before an answer indicates that the answer is synonymous with that which preceded it. Answers separated by a forward slash cannot therefore be taken as different answers.

3. A double forward slash // is used to indicate where multiple answers are given but not all are required.

4. Round brackets ( ) indicate material which is not considered to be essential in order to gain full marks.

5. ‘etc.’ is used in this marking scheme to indicate that other answers may be acceptable. In all other cases, only the answer given or ‘words to that effect’ may be awarded marks.

6. In calculations, 3 marks are deducted for a mathematical error but no further penalty is incurred if the problem, otherwise correct, is completed. Allow for rounding unless specified otherwise; accept an answer given within a reasonable range if the method of calculation is correct.

Current Marking Scheme

Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students’ work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice.

Copyright

All rights reserved. This marking scheme and corresponding papers(s) are protected by Irish (EU) copyright law. Reproduction and distribution of these materials or any portion thereof without the written permission of the publisher is prohibited except for the immediate use within a classroom.

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2018.1 L.35/36_MS 4/60 of 59 examsDEB


Physics

Ordinary Level Marking Scheme (400 marks)

SECTION A (120 marks)

Answer three questions from this section. Each question carries 40 marks.

Section A Question 1 (40 marks)

1. An experiment was set up to measure the acceleration of a trolley. Two velocity measurements were taken for the trolley, along with another measurement. These measurements were used to find the acceleration of the trolley.

(i) Draw a labelled diagram of the apparatus used in this experiment. Indicate on your diagram how the trolley was accelerated. (12)

** Diagram to include: (4 × 3m)

– trolley / glider – runway / airtrack – ticker timer / light gates and timer – method of applying acceleration / tilt track, weight attached to object via pulley // etc.

** Accept any valid alternatives, e.g. data logging. ** Deduct (2m) if no labels included.

(ii) How was the velocity measured? (2 × 3m) (6)

– measure distance between dots / divide length of card – divide distance by time / by time measured by timer

(iii) How was the effect of friction reduced in this experiment? (6)

Any 1: (6m) – clean runway // – oil wheels // – cushion of air in airtrack

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2018.1 L.35/36_MS 5/60 of 59 examsDEB

Section A Question 1 (cont’d.)

(iv) What other measurement needed to be taken in order to calculate the acceleration? (6m) (6)

– time between dots where initial and final velocity were recorded / distance between two light gates

(v) How were the measurements used to calculate the acceleration of the trolley? (10)

Any 1: (10m)

– (acceleration =)12

12

tt

vv

/ t

uva

– asuv 222

** Award 7m for formula with one error. ** Award 4m for any partial answer.

2018 LC Physics [OL]

2018.1 L.35/36_MS 6/60 of 59 examsDEB


2. An experiment was set up to measure the specific latent heat of vaporisation of water. The student took a number of measurements and then added steam to water in a calorimeter. After the steam was added, more measurements were taken.

(i) Draw a labelled diagram of the apparatus used in this experiment. (12)


– water and insulated container – thermometer – steam generator – delivery tube / steam trap

** Deduct 2m if no labels included.

(ii) State three measurements that were taken before the steam was added to the water. (3 × 3m) (9)

– mass of calorimeter – mass of calorimeter + water – temperature of water

(iii) State two measurements that were taken after the steam was added to the water. (2 × 3m) (6)

– mass of calorimeter + water + steam – temperature of water

(iv) How was the mass of the steam calculated? (3 × 3m) (9)

– mass of calorimeter + water + steam – minus – mass of calorimeter + water before adding steam

(v) How could the heat loss of the water have been avoided in the experiment? (4)

Any 1: (4m) – insulation around calorimeter // – starting temperature of water below room temperature


2018.1 L.35/36_MS 7/60 of 59 examsDEB


3. An experiment was set up to measure the focal length of a convex lens. Firstly an approximate value for the focal length was measured. Then a more accurate value was obtained by measuring the distance v of the real image formed from the lens when an object was placed a distance u from the lens. This was repeated at a number of different object distances.

The table shows the data recorded during the experiment.

u (cm) 30 40 50

v (cm) 61 41 34

(i) Draw a labelled diagram of the apparatus used in this experiment. (12)


– illuminated source / object pin – lens – screen / image pin – correct arrangement of equipment


(ii) How was the approximate value of the focal length found? (2 × 3m) (6)

– focus distant object onto screen using lens – distance from lens to screen is approximate focal length

(iii) Calculate the focal length f of the convex lens, using the data above. (15)

– vuf

111 (3m)

Any 1 correct (6m) Other 2 correct (3m)

– 61

1

30

11 f

, f = 20.11 cm

– 41

1

40

11 f

, f = 20.25 cm

– 34

1

50

11 f

, f = 20.24 cm

– Average f = 20.2 cm (3m)

(iv) Why would the student not use an object distance of 10 cm in this experiment? (7)

– 10 cm inside focal point (3m) – virtual image is formed / image can’t

be captured on screen (4m)


2018.1 L.35/36_MS 8/60 of 59 examsDEB


4. An experiment was set up to investigate the relationship between current I and potential difference V for a copper sulfate solution.

The table shows the data recorded during the experiment.

V (V) 1.0 1.5 2.0 2.5 3.0 3.5 4.0

I (A) 0.65 0.98 1.31 1.64 1.97 2.29 2.61

(i) Draw the circuit diagram used to collect the above data. (15)


– battery / power supply – method of varying current – ammeter in series – voltmeter in parallel – copper electrodes in copper sulfate solution


(ii) Using the data in the table, draw a graph, on graph paper, to show the variation of current with potential difference. (12)

** Graph to include: (4 × 3m)

– axes labelled correctly, I against V – 3 points plotted correctly – another 2 points plotted correctly – straight line through the points and origin

** Deduct 3m if graph paper not used.

0.70

0.35

1.05

1.40

1.75

2.10

2.45

2.80

0.00

I (A)

V (V)

1.00.50.0 1.5 2.0 2.5 3.0 4.03.5


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(iii) Calculate the slope of your graph and hence determine the resistance of the copper sulfate solution. (9)

– use slope formula to measure slope, 12

12

xx

yy

e.g. (1.25, 0.8), (2.75, 1.8) (3m)

66.05.1

1

25.175.2

8.08.1

(3m)

– resistance = slope

1=

66.0

1 = 1.52 (3m)

** Accept any reasonable calculation of slope based on graph.

(iv) Use your graph to find the current when the potential difference is 2.2 V. (4)

– line drawn across from 2.2 V on graph (2m) – current =1.40 (2m)

** Accept any reasonable reading from the graph. ** Award full marks for any reading in range 1.39 - 1.41 A.

0.70

0.35

1.05

1.40

1.75

2.10

2.45

2.80

0.00

I (A)

V (V)

1.00.50.0 1.5 2.0 2.5 3.0 4.03.5

I at 2.2 V


2018.1 L.35/36_MS 10/60 of 59 examsDEB

SECTION B (280 marks)

Answer five questions from this section. Each question carries 56 marks.

Section B Question 5 (56 marks)

5. Answer any eight of the following parts (a), (b), (c), etc. ** Marks awarded for the eight best answers.

(a) State the principle of conservation of momentum. (4m + 3m) (7)

– the total momentum before interaction / collision – is equal to the total momentum after, assuming no external forces act on the system / in a

closed system

(b) What is the potential energy gained by a ball of mass 500 g after it has been thrown through a vertical height of 12 m? (7)

– potential energy, PE = mgh (3m) – = (0.5)(9.8)(12) (2m) – = 58.8 J (2m)

(c) What does the U-value of a structure allow you to determine? (4m + 3m) (7)

– how good an insulator a substance is – the lower the U-value, the better the insulation

(d) What change is observed in the sound wave of an ambulance siren by a person as the ambulance passes them on the side of a road? (4m + 3m) (7)

– frequency changes from being higher than emitted frequency – to lower than emitted frequency

** Accept ‘frequency changes from higher to lower’ for full marks.

(e) What is the critical angle of a substance? (4m + 3m) (7)

– the angle of incidence – when the angle of refraction is 90°


2018.1 L.35/36_MS 11/60 of 59 examsDEB

Section B Question 5 (cont’d.)

(f) Name two types of electromagnetic radiation. (7)

Any 2: (4m + 3m) – radio waves // – microwaves // – infra red (IR) // – visible light // – ultraviolet (UV) // – X-rays // – gamma rays

(g) A light emitting diode (LED) contains a p-n junction, which is made of p-type and n-type semiconductors. What is the difference between a p-type and an n-type semiconductor? (4m + 3m) (7)

** Expect one point of information on each type.

p-type

Any 1: – majority charge carriers are positive holes // – short one electron in outer shell // – has an excess of positive holes // – is doped with boron

n-type

Any 1: – majority charge carriers are electrons // – has extra electron in outer shell // – has an excess of electrons // – is doped with phosphorus

(h) Choose from the list below the scientist after which the units of (i) electrical current and (ii) magnetic flux density are named. (4m + 3m) (7)

Volta Ampere Tesla Faraday

(i) electrical current – Ampere (ii) magnetic flux density – Tesla

(i) What is the energy of a photon of frequency 5 × 1015 Hz? (7)

– E = hf (2m) – (6.6 × 10–34)(5 × 1015) (3m) – = 3.3 × 10–18 J (2m)


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(j) State what the three symbols represent in Einstein’s formula E = mc2. (7m) (7)

** Any 1 correct (3m). ** Any 2 correct (5m). ** All 3 correct (7m).

E – energy m – mass c – speed of light


2018.1 L.35/36_MS 13/60 of 59 examsDEB


6. (i) What is meant by the moment of a force? (2 × 3m) (6)

– force multiplied by – distance from fulcrum

** Award full marks for correct notation explained, moment = F × d. ** Award 3m for ‘turning effect of a force’.

(ii) What is a lever? (2 × 3m) (6)

– an object / rigid body free to rotate about – a fixed point called a fulcrum

(iii) In terms of turning forces, explain what is meant by a couple? (3 × 3m) (9)

– two equal forces – opposite directions – forces not in line

** Award full marks for correct explanation stated or implied by diagram.

(iv) Where are the positions of the fulcrums when the person is rowing the boat shown above? (6m) (6)

– at the points on the edge of the boat that the oars pass through / oarlocks

(v) Explain why it is an advantage to have longer oars on a row boat. (2 × 3m) (6)

– greater turning effect / moment – for greater distance from fulcrum

or

– larger moment of a force – due to larger distance from the fulcrum / from end of oar


2018.1 L.35/36_MS 14/60 of 59 examsDEB


A long piece of timber is placed over a stand as shown below. A man of mass 80 kg sits on the left-hand side, 60 cm from the stand. A boy of mass 45 kg sits on the right-hand side. The piece of timber remains balanced.

(vi) Calculate the weight of (a) the man and (b) the boy. (3m + 2m) (5)

W = mg

– weight of man = (80)(9.8) = 784 N – weight of boy = (45)(9.8) = 441 N

(vii) At what distance from the stand does the boy sit? (9)

– 784 × 0.6 = 441 × x (2 × 3m) – x = 1.067 m (3m)

Note: as g cancels also accept: – 80 × 60 = 45 × x – x = 106.66 cm x = 1.066 m

(viii) If a girl of mass 35 kg then sits on the piece of timber 30 cm from the stand on the right-hand side, to what distance from the stand would the boy have to move for the piece of timber to be balanced again? (9)

– (784 × 0.6) = (35 × 9.8 × 0.3) + (441 × x) (2 × 3m) (for girl and boy on right-hand side)

– x = 0.833 m (3m)

Note: as g cancels also accept: – 80 × 60 = 35 × 30 + 45 × x – 4800 = 1050 + 45x 3750 = 45x x = 83.3 cm x = 0.833 m


2018.1 L.35/36_MS 15/60 of 59 examsDEB


7. The sound waves produced by the musicians below are caused by vibrations, and for both the cello and the tin whistle, stationary waves are set up. In the tin whistle, the stationary wave is set up on the pipe with both ends being open.

(i) When producing sound, what vibrates in a cello and in a tin whistle? (2 × 3m) (6)

- cello – strings - tin whistle – (column of) air

(ii) What is a stationary wave? (2 × 3m) (18)

– wave produced when two waves travel in opposite directions / wave where there is no movement along direction of waves / wave that has positions where there is no amplitude called nodes

– causing interference / only oscillations perpendicular to direction of wave / and positions of maximum amplitude called antinodes

or

– waves moving in opposite directions interfere with each other – resulting in no net movement of wave

Draw a diagram of a stationary wave set up (a) on a string in the cello and (b) in the tin whistle when they are vibrating at fundamental frequency.

(a) string in cello (2 × 3m)

– two nodes at each end – antinode at middle

(b) tin whistle (2 × 3m)

– antinode open at both ends – node in the centre

(iii) What are the names given to the positions of maximum and minimum vibrations? (2 × 3m) (6)

- maximum – antinode - minimum – node


Node Node

Antinode

NodeAntinode Antinode

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(iv) While playing the cello, how does the musician moving his finger up and down the string result in a different sound being produced? (2 × 3m) (12)

– moving finger changes (effective) length of string – changes frequency

If he moves his finger along the string from the top of the cello to the middle, what change occurs to the sound? (2 × 3m)

– shortening of length of string / as f l

1

– causes frequency to increase

(v) If the cello string vibrates at 282 Hz, when the distance between the fixed ends of the string is 0.55 m, calculate the speed of the wave in the string. (8)

– = 2(0.55) = 1.1 m (3m) – c = f = (282)(1.1) (3m) – c = 310.2 m s–1 (2m)

(vi) Why does the tin whistle have a number of holes along its length? (2 × 3m) (6)

– blocking / unblocking holes varies the length of the pipe – varies frequency of sound produced


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8. (i) Define potential difference. (2 × 3m) (6)

– work done – per unit change / in moving a charge of 1 C from one point to another

** Award full marks for correct notation explained, V = Q

W.

(ii) What is meant by the capacitance of a conductor? (2 × 3m) (6)

– the ratio of charge – to potential difference

** Award full marks for correct notation explained, C = V

Q.

(iii) How does the capacitance of an isolated conductor (i.e. no loss or gain of charge) change as its potential is increased? (2 × 3m) (6)

– as C = V

Q, capacitance decreases

– as potential increases

(iv) Describe an experiment to show that capacitors store energy. (4 × 3m) (12)

– connect battery across capacitor until fully charged – disconnect battery – connect capacitor to bulb – bulb lights for short period of time

A circuit is set up with a parallel plate capacitor connected to a 24 V d.c. power supply as shown.

(v) What charge is on the capacitor when it is fully charged? (9)

– C = V

Q (3m)

– Q = (2.6 × 10–6)(24) (2 × 2m) for each correct substitution – Q = 6.24 × 10-5 C (2m)

(vi) Redraw the circuit and indicate on your diagram where the charge forms on the capacitor. (2 × 3m) (6)

– positive charge shown on the side of the capacitor connected to the positive terminal – negative charge shown on the side of the capacitor connected to the negative terminal


24 V

2.6 μF

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(vii) A short time after connecting the battery to the capacitor, what current is observed? (3m) (6)

– current is zero / no current

What difference would be observed if an a.c. power supply was connected across the capacitor? (3m)

– current continually observed

(viii) Give two uses for a capacitor. (5)

Any 2: (3m + 2m) – tuning radios // – flash guns // – smoothing currents in rectifier // – filtering frequencies // – blocking direct current (d.c.) // etc.

** Accept any appropriate answer(s).


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9. (i) State Faraday’s law of electromagnetic induction. (3 × 3m) (9)

– electromotive force (emf) / voltage induced / produced – is proportional to rate of change – of magnetic flux

A magnet is moved towards a coil in such a way that the south pole faces the coil as shown.

SN S

N

(ii) Explain why a south pole is produced at the side of the coil nearest the magnet. (3 × 3m) (9)

– by Lenz’s law, current induced produces pole that opposes – change that caused it, i.e. south pole moving towards coil – south pole induced in coil opposes south pole of magnet

(iii) If the magnet is then pulled away from the coil, as shown below, why is a north pole produced at the end nearest the magnet? (2 × 3m) (6)

SN

S

N

– current induced produces north pole – to prevent magnet moving away

(iv) What is observed when the magnet is stationary inside the coil? (3m) (6)

– no current / deflection of galvanometer

Explain your answer. (3m)

– no changing magnetic field to induce electromotive force (emf)


2018.1 L.35/36_MS 20/60 of 59 examsDEB


An a.c. source is connected to the primary coil shown.

Inputcoil

Outputcoil

Primary Secondary

(v) Why does a current flow in the secondary coil? (3 × 2m) (6)

– changing current in primary coil produces changing magnetic field – changing magnetic field cutting secondary coil induces an electromotive force (emf) – emf causes current to flow in secondary coil

(vi) Draw a sketch of the output of the secondary coil. (6)

– graph of voltage varying with time (3m) – sine wave shape (3m)

(vii) There are 200 turns in the primary coil and 12 turns in the secondary coil. Calculate the output voltage when 230 V is connected across the primary coil. (8)

– p

s

p

s

N

N

V

V (3m)

– 200

12

230s V

(3m)

– Vs = 13.8 V (2m)

(viii) The resistance of the secondary coil is 6 Ω. Calculate the current flowing in the secondary coil. (6)

V = IR – 13.8 = I(6) (3m) – 2.3 A (3m)


2018.1 L.35/36_MS 21/60 of 59 examsDEB


10. (i) State three properties of electrons. (9)

Any 3: (3 × 3m) – negatively charged // – smallest subatomic particle // – orbit the nucleus // – deflected in magnetic and electric fields // – cause fluorescence

(ii) What is the difference between thermionic emission and photoelectric emission? (4 × 3m) (12)

– emission of electrons – from surface of a metal – thermionic – due to heat – photoelectric – due to light (electromagnetic radiation)

In an experiment to investigate the photoelectric effect, a zinc plate was placed on the cap of a gold leaf electroscope as shown. The gold leaf electroscope had been negatively charged.

−−− −

−−−

Gold leaves

Electroscope

CapZinc plate

Radiaon

(iii) Describe how the negative charge was placed on the electroscope. (4 × 3m) (12)

– positively charged rod brought near the cap – negative charge attracted to cap, positive charge repelled to leaves – ground cap → positive charge repelled to ground, negative charge held at cap – remove grounding and positive rod, negative charge spreads to gold leaves

(iv) What was observed when UV light was shone on the zinc plate? (3m) (6)

– gold leaves converge / fall


– electrons are removed by UV light


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(v) What was observed when a weak beam of red light was incident on the zinc plate? (3m) (11)

– no change to gold leaves

How did this change when an intense beam of red light was incident on the zinc plate? (2m)

– no change to gold leaves

** Accept ‘in both cases no change to gold leaves’ for (5m).

Explain these observations. (2 × 3m)

– frequency of red light too low – to remove electrons

(vi) Give two applications of the photoelectric effect. (6)

Any 2: (2 × 3m) – photocell // – burglar alarms // – photocopiers // – photodiodes // – automatic doors // – optical soundtrack in films // – safety switches on cutting machinery // – control sensors in central heating boilers // – laboratory light meters // etc.

** Accept any appropriate answers.


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11. Read the following passage and answer the questions below.

(a) Define pressure. (4m + 3m) (7)

– force divided by – area

** Award full marks for correct notation explained, P = A

F.

(b) Why is the pressure acting on a person much greater at depth under water than on the surface of the Earth? (7)

Any 1: (7m) – water has a much higher density than air // – much bigger weight of water above person at depth

(c) How many atmospheres of pressure does a diver feel at a depth of 50 m? (4m + 3m) (7)

– 10

50

– = 5 atmospheres


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(d) Why was the Styrofoam cup much smaller after being lowered deep under water? (4m + 3m) (7)

– much higher pressure – squeezes air out of the Styrofoam

(e) State Boyle’s law. (4m + 3m) (7)

– pressure is inversely – proportional to volume (for a fixed mass of gas at a constant temperature)

** Award full marks for correct notation explained, P V

1 .

(f) Calculate the pressure (in pascals) at a depth of 10 m. (7)

– ghP (2m)

– = (1000)(9.8)(10) (3m) – = 98000 Pa (2m)

(g) A diver can experience ‘the bends’ if he or she comes back up to the surface too quickly. Describe what happens when a diver experiences ‘the bends’. (4m + 3m) (7)

– at great depth, too much nitrogen is dissolved in the blood – returning to surface too quickly results in nitrogen forming bubbles in blood

(h) Describe the type of weather we get under conditions of high atmospheric pressure. (7)

Any 2: (4m + 3m) – dry // – good // – clear skies // – little wind / settled / calm

(acceleration due to gravity, g = 9.8 m s–2; density of water = 1000 kg m–3)


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Section B Question 12 (2 × 28 marks)

12. Answer any two of the following parts (a), (b), (c), (d).

(a) (i) Define velocity. (6)

Any 1: (2 × 3m)

– speed in a – given direction

or

– change in displacement – divided by time

** Award full marks for correct notation explained, v = t

s.

A train starts from rest at a station and accelerates at 1.2 m s–2 for 60 seconds. It travels at this speed for 5 minutes. It then decelerates to rest in 40 seconds.

(ii) Calculate the maximum velocity reached by the train. (6)

– atuv (3m) – = 0 + (1.2)(60) = 72 m s–1 (3m)

(iii) Calculate the distance travelled by the train during deceleration. (8)

– 0 = 72 + a(40), a = –1.8 m s–1 (4m)

– 2

2

1atuts , (72)(40) + ½(–1.8)(40)2 (3m)

– = 1440 m (1m)

or

– tvu

s

2 (4m)

– 402

072

s (3m)

– = 1440 m (1m)

(iv) Sketch a velocity–time graph of the train journey. (8)

** Graph showing:

– correct acceleration part of graph (3m) – correct constant velocity part of graph (2m) – correct deceleration part of graph (3m)


2018.1 L.35/36_MS 26/60 of 59 examsDEB


(b) In 1911, Ernest Rutherford performed an experiment where he fired beams of alpha particles at a thin gold foil. The directions of the beams after passing through the gold foil were detected on a fluorescent screen.

Circular fluorescentscreen

Thin goldfoil

Source of α parcles

Parcle beam

(i) How are alpha particles produced? (2 × 3m) (6)

– emitted from / produced by spontaneous decay of – radioactive nuclei

(ii) Describe what Rutherford observed in this experiment. (3 × 3m) (9)

– most alpha particles went through undeflected – some deflected through a small angle – a very small number deflected backwards / through > 90°

(iii) What conclusion did Rutherford make about the structure of an atom based on these observations? (2 × 3m) (6)

– all the positive charge – is concentrated in the small volume at the centre of the atom (the nucleus)

** Accept ‘proposed existence of the nucleus’ for 6m.

(iv) Niels Bohr proposed a model in 1913 that furthered the understanding of the structure of the atom. What did his theory propose? (4m + 3m) (7)

– electrons orbit around the nucleus – electrons are restricted to specific orbits


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(c) (i) Distinguish between heat and temperature. (2 × 3m) (6)

– heat is a form of energy – temperature is a measure of hotness

Water in a beaker is heated from room temperature at 20 °C until it is boiling.

(ii) What is the maximum and minimum temperature of the water on the Kelvin scale to the nearest Kelvin? (3 × 3m) (9)

– T (K) = 273(.15) + T (°C) → stated or implied – maximum - 100 °C = 373 K – minimum - 20 °C = 293 K

A thermistor is placed in the water as it is being heated. The thermistor is connected to an instrument that measures a thermometric property associated with the thermistor.

(iii) Identify the instrument and the thermometric property that it measures. (6)

Instrument (3m)

– ohmmeter / multimeter

Thermometric property measured (3m)

– resistance

(iv) Why would a standard thermometer also need to be placed in the water so that the thermistor could be used as a thermometer? (7)

Any 1: (4m) – temperatures of the water would not be known at resistances other than

for boiling water // – different thermometric properties would vary differently with temperature

Name a suitable standard thermometer. (3m)

– mercury-in-glass thermometer

** Accept any valid commercial laboratory thermometer.


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(d) (i) What is diffraction? (2 × 3m) (6)

– the spreading out of a wave – when it is incident on a gap or obstacle

When a monochromatic laser beam is incident on a diffraction grating, an interference pattern is formed.

Laser n = 0

n = 1

n = 2

n = 2

n = 1

(ii) What is monochromatic light? (4m) (4)

– (a monochromatic light source) produces a single wavelength / frequency / colour

(iii) Why do bright images appear only at certain positions on the wall? (4m) (4)

– constructive interference occurs / crests meets crests

(iv) If the grating constant is 4 × 10–6 m and the angle between the n = 0 and n = 2 images is 18.4°, calculate the wavelength of the monochromatic laser beam. (8)

– sindn (2m) – 2 = (4 × 10–6)(sin 18.4) (3m) – = 6.31 × 10–7 m (3m)

(v) What would you observe if a source of white light is used instead of the monochromatic light source? (2 × 3m) (6)

– spectra of light / number of different colours // – at each order / n


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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 verify the principle of conservation of momentum, a body A was set in motion at constant velocity and collided with a second body B, initially at rest. The two bodies stayed together after the collision. The velocities before (v1) and after (v2) the collision were calculated using measurements taken in the experiment. The measurements were repeated for different masses of A and B.

The following data were recorded.

Mass of A (g) Mass of B (g) v1 (m s–1) v2 (m s–1)

200 150 0.845 0.484

250 300 0.632 0.288

Draw a labelled diagram of the apparatus used. (9)

Diagram showing: (3 × 3m)

– track, e.g. (airtrack or runway) – two trolleys and method of coalescing, e.g. Blu-tack, magnets, velcro – labelled method of measuring velocity or time (motion sensor, ticker-tape timer,

light gates and timer)

examsDEB

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State what measurements were taken to calculate the velocities in this experiment and explain how these measurements were used to calculate the velocities. (9)

Any 1: (3 × 3m)

Ticker-tape timer

– time between dots = 1/50 s – measure distance (d) between n dots

– speed =

501

n

d

Light gates and timer

– time card passes through light gate measured by timer (t) – measure length (l) of card

– speed = t

l

Motion sensor

– distance time graph produced by sensor – take two points from graph – speed = slope of graph

Use the data from the table to show how this experiment verifies the principle of conservation of momentum. (12)

– 2BA1A )( VmmVm (3m)

– (0.2)(0.845) = (0.35)(0.484) (3m) – 0.169 ~ 0.1694 kg m s–1 - verifying conservation of momentum (3m) (0.25)(0.632) = (0.55)(0.288) – 0.158 ~ 0.1584 kg m s–1 - verifying conservation of momentum (3m)

** Deduct 1m for omission of or incorrect unit.

To ensure momentum is conserved, the effects of external forces acting on the bodies needs to be minimised.

Identify two external forces that may be present in the experiment and state how the effect of one of these forces may be minimised. (10)

External forces (2 × 3m)

– friction – gravity

How minimise effect of an external force

Any 1: (4m) - friction – can be removed by air cushion on airtrack // – oil wheels on trolleys // – remove dirt on track

- gravity – adjust slope of runway or track for constant velocity // – if track is horizontal there is no external gravity force

2018 LC Physics [HL]

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2. The specific heat capacity of water was found by adding hot copper to water in an insulated aluminium calorimeter. The following data were recorded.

Mass of calorimeter = 56.4 g Mass of calorimeter + water = 111.6 g Mass of copper = 84.1 g Initial temperature of water = 16.2 °C Temperature of hot copper = 99 °C Final temperature of water = 24.6 °C Room temperature = 20.6 °C

Explain how the temperature of the hot copper was measured. (3m) (6)

– thermometer / place in boiling water / temperature sensor

Give a reason why the correct reading for the temperature of the copper added to the water may be less than this measurement.

Any 1: (3m) – temperature would have dropped from measurement to when placed in water // – low specific heat capacity of copper means temperature drops quickly

Why should the initial temperature of the water be set below room temperature at the start of the experiment? (3 × 3m) (9)

– heat lost to surroundings while above room temperature – is balanced by – heat gained from surroundings while water is below room temperature

Calculate the specific heat capacity of water. (15)

– heat lost by copper = heat gained by water + calorimeter (stated or implied) (3m) – 2aa2ww1cc cmcmcm (3m)

– (0.0841)(390)(99 – 24.6) = (0.1116 – 0.0564)cw(24.6 – 16.2) + (0.0564)(921)(24.6 – 16.2) (3m) – 2440.2456 = 0.46368cw + 436.33 (3m) – 4321.8 J kg–1 K–1 (3m)


The correct value for the specific heat capacity of water is 4180 J kg–1 K–1. Assuming that the error in the value calculated is due to the fact that the temperature of the copper added to the water was less than the temperature of the hot copper measured, calculate the actual temperature of the copper added to the water. (10)

– (0.0841)(390)(T – 24.6) = (0.0552)(4180)(24.6 – 16.2) + (0.0564)(921)(24.6 – 16.2) (3m) – 32.799(T – 24.6) = 2374.515 (3m) – T – 24.6 = 72.4 (2m) – T = 97.0 °C (2m)


(specific heat capacity of copper = 390 J kg–1 K–1; specific heat capacity of aluminium = 921 J kg–1 K–1 )


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3. In an experiment to measure the focal length of a concave mirror, a student recorded the following values of the image distance v for a number of different object distances u.

u (cm) 30.0 35.0 40.0 45.0 50.0 55.0

v (cm) 70.4 52.5 44.2 39.4 36.5 34.2

Describe, with the aid of a labelled diagram, how the image distance v was measured. (12)

Diagram showing: (6m)

– object – correct mirror – screen / search pin

** Award (3m) for any two correct items.

Description (2 × 3m)

– adjust position of screen until sharp image is obtained / adjust search pin position until no parallax – measure distance from screen / search pin to centre of mirror

Calculate the focal length of the concave mirror by drawing a suitable graph based on the data given above. (18)

Any 1:

Graph of u

1 against

v

1

u

1 (cm–1) 0.033 0.0286 0.0250 0.0222 0.0200 0.0182

v

1 (cm–1) 0.0142 0.0190 0.0226 0.0254 0.0274 0.0292

– calculate u

1 against

v

1values (3m)

– label axes (3m) – plot points correctly (3m) – straight line (3m)

– read off points where line cuts axes f

1 (range 0.046-0.049) (3m)

– average f = 21.0 ± 0.7 (3m)


or


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Graph of u against v

– label axes (3m) – plot points correctly (3m) – curve through points (3m) – read off point where u = v (3m) – range 41-43 cm (3m) – f = 0.5 (41 to 43) = 21.0 ± 0.5 (3m)

** Deduct 1m for omission of or incorrect unit. ** Award a maximum of 12m if f is calculated using formula without using graph.

Why did the student not choose an object distance u of 20 cm during this experiment? (2 × 3m) (6)

– 20 cm is inside focal point – a virtual image is produced image can’t be captured on screen

Why are the values measured for u generally more accurate than the values measured for v? (4)

Any 1: (4m) – position of object is known exactly // – there is uncertainty in position of sharpest image


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4. The following is part of a student’s report on an investigation of the variation of current I with potential difference V for a semiconductor diode.

“I set up the circuit with the diode in forward bias. I measured the current through the diode for different values of the potential difference across the diode.”

The following data were recorded.

V (V) 0 0.25 0.50 0.60 0.62 0.64 0.66 0.68 0.70 0.72 0.75

I (mA) 0 0.5 0.9 2.6 6.3 11.2 19.4 29.1 45.2 65.3 96.7

Draw the circuit diagram used by the student to obtain these data. (9)

Diagram showing:

– power supply, including method of varying voltage (3m) – voltmeter and ammeter in correct arrangement (3m) – diode in forward bias (3m)

–

** Do not penalise if amounts and units for voltage (9V) and resistance (300 ) are not included in diagram.

Draw a graph showing how the current varies with the potential difference for the semiconductor diode. (9)

Graph

– axes labelled correctly (3m) – points plotted correctly (3m) – curve through points (3m)


mA300 Ω

9 VV

V (V)

I (mA)

0.30 0.50 0.700.600.400.200.10 0.80

0.000.0

20.0

60.0

70.0

80.0

90.0

100.0

30.0

40.0

50.0

10.0

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Explain why the current varies with potential difference as shown by your graph. (3 × 3m) (9)

– current very low up to approximately 0.6 V – junction voltage (~0.6 V) has not been overcome – when applied voltage is above junction voltage, current increases rapidly

The student’s report continued…

“I then changed the circuit so that the diode was set up in reverse bias and once again measured the current through the diode for different values of the potential difference across the diode.”

What changes did the student make to the circuit? (3 × 3m) (13)

Stated or shown in diagram

– direction of diode reversed – milliammeter replaced by microammeter – voltmeter placed across diode and microammeter

–

** Do not penalise if amount and unit for voltage (9V) are not included in diagram.

How would the graph obtained for current against potential difference for the diode in reverse bias differ from that obtained for the diode in forward bias? (4m)

– zero or very small negative current

** Accept a sketch of graph for full marks.

–

–

μA9 V

V


V (V)

(mA)I

V (V)

(μA)I

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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) What are the two conditions for a body to be in a state of equilibrium? (4m + 3m) (7)

– sum of forces in any direction is zero / forces up = forces down – sum of moments about any point is zero / clockwise moments = anticlockwise moments

(b) A trolley is dragged with a force of 200 N at an angle of 30°. What is the work done on the trolley when it travels a distance of 50 m? (4m + 3m) (7)

– force in direction of motion = 200 cos 30 = 173.2 N – W = F × d = 173.2 × 50 = 8600 J


(c) Draw a ray diagram showing how a prism can turn a ray of light through 180°. (4m + 3m) (7)

Ray diagram showing:

– light striking two internal interfaces of prism at 45° – total internal reflection at both surfaces

(d) Explain the principle that causes the inside of a refrigerator to become cold. (4m + 3m) (7)

– change of state of circulating liquid to a gas – requires latent heat to be extracted from inside of fridge

(e) What is the frequency emitted by a train whistle if it is observed by a person standing on the platform of the station as being 940 Hz when the train is leaving the station at 40 m s–1? (4m + 3m) (7)

uc

fcf

'

– 940 = 40340

)340(

f

– f = 1050.6 Hz



30˚

200 N


45˚

45˚

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(f) State two adjustments that need to be made to a spectrometer before it is used. (7)

Any 2: (4m + 3m) – cross threads in focus // – focus telescope on distant object // – adjust collimator for parallel light // – adjust width of slit // – level turntable

(g) How does the resistance of a conductor vary with length and cross-sectional area? (4m + 3m) (7)

– (resistance is) proportional to length – (resistance is) inversely proportional to cross-sectional area

(h) Calculate the energy stored on a parallel plate capacitor that has a charge of 10 μC when connected across a potential difference of 15 V. (4m + 3m) (7)

– V

QC =

15

)1010( 6 = 6.66 × 10–7 F

– W = ½ CV2 = ½ (6.6 × 10–7)(15)2 = 7.4 × 10–5 J


(i) Give an example of an equation for nuclear fusion of hydrogen. (7)

Any 1: (7m)

– n10

32

21

21 HeHH //

– n10

42

31

21 HeHH //

– HeHH 42

21

21

** Award 2m + 2m for correct elements on each side of the equation. ** Award 3m for correct atomic and mass numbers.


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(j) Name the scientist who (i) predicted mathematically the existence of antiparticles, (ii) won Ireland’s only Nobel prize for physics. (4m + 3m) (7)

(i) predicted mathematically the existence of antiparticles

– Paul Dirac

(ii) won Ireland’s only Nobel prize for physics

– Ernest Walton

or

State the principle on which the moving coil loud-speaker is based. (4m + 3m) (7)

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

(speed of sound in air = 340 m s–1

)


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6. State Newton’s third law of motion. (6)

Any 1: (2 × 3m)

– every action has an equal – and opposite reaction

or

– force exerted by A on B is equal – and opposite to force exerted by B on A

Explain how F = ma can be derived from Newton’s second law of motion. (9)

– F is proportional to t

mumv (2m)

– F is proportional to ma (2m) – F = kma (2m) – k = 1 by definition of the newton. F = ma (3m)

An elevator starts from rest and accelerates upwards at 2 m s–2 for 3 seconds. It then travels at constant speed for 10 seconds and finally decelerates to rest in 2 seconds.

Calculate the constant speed that the elevator reaches. (2 × 3m) (15)

– atuv – = 0 + (2)(3) = 6 m s–1


Draw an accurate velocity-time graph representing the motion of the elevator.

Velocity-time graph showing

– correct acceleration part of graph (2m) – correct constant velocity (2m) – correct deceleration part of graph (2m) – correct labelling and scale on graph (3m)

–


13 t (s)

v (m s–1)

153

6

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Calculate the distance travelled by the elevator during its ascent. (12)

Any 1: (4 × 3m)

– 2

2

1atuts = (0)(3) + ½(2)(3)2 = 9 m // area = ½(3)(6) = 9 m

– s = (6)(10) = 60 m // area = (6)(10) = 60 m – v = 6 – a(2), a = –3, s = (6)(2) + ½(–3)(2)2 = 6 m // area = ½(2)(6) = 6m – total distance = 9 + 60 + 6 = 75 m


Calculate the maximum force exerted by the floor of the elevator on a person of mass 80 kg in the elevator as it accelerates upwards. (8)

When accelerating

– net force = N – mg = ma (3m) – N = ma + mg = (80)(2) + (80)(9.8) (3m) – = 944 N (2m)


If the person experienced weightlessness during the descent of the elevator, what would be the acceleration of the elevator? (3m) (6)

– acceleration = 9.8 m s–2


– normal reaction = 0 / no contact force between person and elevator / person is in freefall

(acceleration due to gravity, g = 9.8 m s–2)


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7. Define electric field strength and give its unit of measurement. (17)

electric field strength (2 × 3m)

– force per – unit charge

** Award full marks for correct notation explained, Q

FE .

unit of measurement (3m)

– N C–1 / V m–1

Write an expression for the electric field strength at a point a distance d from the centre of an object containing a charge Q.

– 2

21

4 d

QQF

(3m)

– 2

1

4

)1(

d

QE

(3m)

– 2

1

4 d

QE

(2m)

** Award full marks for final answer written directly.

Describe how an isolated metal sphere can be positively charged by induction. (4 × 3m) (12)

– hold negatively charged rod near metal sphere – attracts positive charge to side of rod / repels negative charge to far side of sphere – ground sphere with finger or otherwise, negative charge on sphere goes to ground – remove finger and remove rod, positive charge remains

Two isolated positively charged spheres are placed 50 cm apart in air. One sphere has a charge Q and the other has a charge 2Q. (27)

Find the electric field strength, in terms of Q, at the midpoint between the two charges. (3 × 3m)

– 22 44

2

d

Q

d

QE

– )25.0)(1085.8(4)25.0)(1085.8(4

2212212

QQE

– = 1.439 × 1011 Q


50 cm

Q 2Q

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Find the position where the electric field strength is zero.

– 22

21 44

2

d

Q

d

Q

(3m)

– Q

Q

d

d

221

22

2

1

1

2 d

d (3m)

– 502 22 dd (2m)

– d2 = 20.71 cm from Q or – d1 = 29.29 cm from 2Q (2m)


A block of concrete is then placed between the two spheres and the spheres are moved so that they are both in contact with either side of the block.

Calculate the length of the concrete block if the electric field strength at the midpoint between the spheres is the same as the electric field strength at the midpoint when they were 50 cm apart in air.

– 05.4 (2m)

– 2

02

0 )5.4(4)5.4(4

2

d

Q

d

QE

= 1.439 ×1011 Q (2m)

– 11212

10439.1))1085.8(5.4(4

1

dE (2m)

– d = 0.1178 m, distance between charges = 2d = 0.236 m (2m)


(relative permittivity of concrete = 4.5)


2018.1 L.35/36_MS 43/60 of 59 examsDEB


8. What is a stationary wave? (9)

Any 1: (3m) – the amplitude of the wave at any point is constant // – there is no net transfer of energy

Explain how stationary waves are produced. (2 × 3m)

– waves of same amplitude and frequency – moving in opposite directions meet

The flute and the clarinet are two approximately cylindrical wind instruments that are played in different ways.

The flute is blown from the side (as shown) so that the tube is open at both ends.

The clarinet is blown through a narrow reed at one end (as shown). When playing the clarinet, there is effectively one end closed.

Draw a diagram of the first two harmonics set up in a flute and in a clarinet, indicating which harmonic each diagram represents. (13)

Flute

Correct first harmonic (3m)

–

Correct second harmonic (2m)

–

Clarinet

Correct first harmonic (3m)

–


NodeAntinode Antinode

Node Node

Antinode Antinode Antinode

AntinodeNode

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Clarinet (cont’d.)

Correct ‘second’ harmonic (third harmonic for a closed pipe) (2m)

–

Harmonics labelled correctly (3m)

The variation of the speed of sound with temperature is shown below.

Temperature (°C)

Spee

d of

soun

d (m

s−1)

325

330

335

340

345

350

355

360

365

10 20 30 40 50 600

How does the speed of sound change with temperature? (3m) (6)

– speed of sound varies linearly with temperature

State another factor on which the speed of sounds depends.

Any 1: (3m) – medium // – pressure

In an orchestra, one musician has a flute of length 65 cm and another musician has a clarinet of length 56 cm. At the start of a concert the temperature of the auditorium was 18 °C.

Using the graph, calculate the fundamental frequencies produced by the flute and clarinet when all holes are blocked in both instruments. (10)

– from the graph: 18 °C → 342 m s–1 (2m) – for flute, = 2l = 2(0.65) = 1.3 m (2m)

–

vf =

3.1

342 = 263.1 Hz (2m)

– for clarinet, = 4l = 4(0.56) = 2.24 m (2m)

–

vf =

24.2

342 = 152. 7 Hz (2m)

** Accept any reasonable reading for 18 °C from the graph. ** Deduct 1m for omission of or incorrect unit.


NodeNode

Antinode Antinode

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Temperature (°C)

Spee

d of

soun

d (m

s−1)

325

330

335

340

345

350

355

360

365

10 20 30 40 50 600

349 m s−1

18 °C

Where, in one of the instruments, would a hole have to be unblocked in order to produce the same fundamental frequency in both instruments? (18)

For clarinet

– l

f4

342 = 263.1 (3m)

– l = 0.324 m or 32.4 cm from closed end (3m)


Even if both instruments produce the same fundamental frequency and amplitude, why will they produce different sounds? (4m)

– different harmonics / overtones produced by the two instruments

Later in the concert the fundamental frequency of the flute changed to 268.5 Hz when all holes were blocked. To what had the temperature of the auditorium changed?

– c = 2f l = 2(268.5)(0.65) (3m) – = 349 m s–1 (2m) – corresponds to 31 °C from the graph (3m)

** Accept any reasonable reading for 349 m s–1 from the graph. ** Deduct 1m for omission of or incorrect unit.


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9. State Lenz’s law of electromagnetic induction. (2 × 3m) (15)

– direction of current induced – is such as to oppose the change that caused it

Describe an experiment that demonstrates Lenz’s law. (3 × 3m)

– a moving magnet induces a current – in a non-ferromagnetic metal / aluminium ring – magnet slowed / metal moved by magnetic field induced in coil / ring moves away from

approaching magnet

** Accept any other appropriate answers.

How does Lenz’s law contribute to the explanation of self-induction? (3 × 3m) (9)

– when current passing through a coil changes – changing magnetic field induces an electromotive force (emf) in the coil – by Lenz’s law, the induced emf in the coil opposes the changing current

** Accept other appropriate answers.

A magnet that is moving towards a 20-turn coil of resistance 12 Ω at 0.5 m s–1 results in a current of 0.25 A produced in the coil.

Calculate (15)

(i) the heat dissipated in the coil each second

– heat produced per second = I2R (3m) – (0.25)2(12) = 0.75 J (3m)


(ii) the force exerted by the coil on the magnet.

– distance travelled per second = 0.5 m (3m) – work done on magnet = F × d = 0.5 F (3m) – 0.5 F = 0.75, F = 1.5 N (3m)



2018.1 L.35/36_MS 47/60 of 59 examsDEB


Another coil, with 100 turns and resistance 20 Ω, is placed near the first coil to produce a transformer. (17)

Calculate the voltage and current induced in the second coil, if transfer of energy is 80% efficient.

– primary voltage = (0.25)(12) = 3 V (3m)

– p

s

p

s

N

N

V

V ⇒ 320

100s V = 15 V (3m)

– output power = 0.8(input power) (2m) – ss IV 0.8( ppIV ) ⇒ (15)Is = (0.8)(3)(0.25) (2m)

– Is = 0.04 A (3m)


State one way of making this transformer more efficient.

Any 1: (4m) – coils closer to each other // – coils wound on same core // – core laminated to avoid eddy currents // – thicker wire used in coils


2018.1 L.35/36_MS 48/60 of 59 examsDEB


10. In 1899, less than 3 years after the discovery of radioactivity by Henri Becquerel, Ernest Rutherford experimentally verified that there were at least two types of radiation produced from a uranium sample, based on the penetrating powers of the radiation produced. He named these different types of radiation alpha (α) and beta (β) rays.

What are α and β rays? (2 × 3m) (12)

α rays – fast-moving helium nuclei β rays – high-speed electrons

State a different property of each.

** Expect a different property for each.

α rays

Any 1: (3m) – positively charged // – deflected in electromagnetic or electric fields // – high ionising ability // – low penetrating power

β rays

Any 1: (3m) – negatively charged // – deflected in electromagnetic or electric fields // – medium ionising ability // – medium penetrating power // – any electron property

Describe an experiment that compares the penetrating powers of α and β rays and another type of radiation emitted from the nucleus of a radioactive atom. (4 × 3m) (12)

– method of detecting radiation (Geiger-Müller tube and counter / solid state detector) – detecting each source of radiation – with suitable minimum obstacle capable of blocking – correct order of penetration (stated or implied): α, least penetrating, β, γ, most penetrating

Francium–223 is the second rarest of naturally occurring radioactive elements; only 30 g is estimated to be present in the Earth’s crust. It is formed by alpha decay of actinium. In a sample of uranium, it is estimated that there is 1 francium atom for every 1 × 1018 uranium atoms.

Write the nuclear equations to represent the above decay of uranium–235 to francium–223. (15)

Correct identification of Th, Ra and Ac (2 × 3m) Correct mass and atomic numbers for Th, Ra and Ac (3 × 1m) Correct labelling of decays (3m)

– FrAcRaThU 22387

22789

22788

23190

23592


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Why is francium so rare in nature in comparison to uranium? (3m)

– half-life of francium (22 minutes) much less than half-life of uranium / decays very quickly after forming

How many atoms of francium are produced by alpha decay in 65.31 years from a sample of 100 g of actinium–227? (17)

– 227 g of Ac → 6.022 × 1023 atoms (3m)

– 100 g of Ac → 227

)100)(10022.6( 23 = 2.653 × 1023 atoms (3m)

– 77.21

31.65 = 3 half-lives (

8

1 actinium remains,

8

7 converted to francium) (2m)

– )10653.2(8

7 23 = 2.32 × 1023 atoms (3m)


How many actinium atoms are decaying per second after 65.31 years?

– 60602436577.21

693.02ln

21

T

= 1.009 × 10–9 (3m)

– A = N = (1.009 × 10–9)(8

1(2.653 × 1023)) = 3.346 × 1013 Bq (3m)


(half-life of actinium–227 = 21.77 years)


2018.1 L.35/36_MS 50/60 of 59 examsDEB


11. Read the following passage and answer the accompanying questions.

(a) What is the photoelectric effect? (4m + 3m) (7)

– emission of electrons from surface of a metal – by electromagnetic radiation of a suitable frequency

(b) How does Einstein’s photoelectric law explain Lenard’s discovery that doubling the intensity of light had no effect on the energies of electrons produced? (7)

Any 1: (7m) – energy of each photon dependent on frequency, not intensity – electron can pick up energy from one photon only

(c) What is the minimum energy needed to remove an electron from zinc whose threshold frequency is 8.78 × 1014 Hz? (7)

– Φ = hf0 (2m) – = (6.6 × 10–34)(8.78 × 1014) (3m) – = 5.79 × 10–19 J (2m)



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(d) What is the maximum energy of electrons emitted from zinc if radiation of wavelength 328 nm is incident on a piece of zinc? (7)

–

cf =

)10328(

)103(9

8

= 9.15 × 1014 Hz (3m)

– ½ mVmax2 = hf – hf0 = (6.6 × 10–34)(9.15 × 1014) – (6.6 × 10–34)(8.78 × 1014)

= 2.442 × 10–20 J (4m)


(e) Why are not all electrons emitted with this kinetic energy? (4m + 3m) (7)

– all electrons do not have same binding energy / some electrons need more energy to escape – less energy available for kinetic energy

(f) List four forms of electromagnetic radiation in order of increasing wavelength. (7)

Any 4: (4 × 1m) In order of increasing wavelength (3m) – gamma → X-ray → UV → visible → IR → microwaves → radio waves

(g) Which one of the four forms of electromagnetic radiation referred to in part (f) would be diffracted through the greatest angle when passing through a diffraction grating? (7)

– longest wavelength of those listed (3m)


– refer to n = dsin , greater wavelength, greater angle of diffraction

(h) Explain how electromagnetic radiation causes the temperature of the Earth to increase according to the greenhouse effect. (7)

– electromagnetic radiation that strikes the Earth is reflected at higher wavelength (generally in IR range) (3m)

– increased carbon dioxide levels in atmosphere reflect back infra-red and heat up Earth’s surface (4m)


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Section B Question 12 (2 × 28 marks)

12. Answer any two of the following parts, (a), (b), (c), (d).

(a) What is centripetal force? (2 × 3m) (6)

– force on an object travelling in a circle – directed towards the centre of the circle

Io is the fourth-largest moon of Jupiter, with an orbit of radius 421,700 km around Jupiter. The period of Io’s orbit is 42.46 hours.

Io

GanymedeEuropa

Callisto

Jupiter

What is the angular velocity of Io during its orbit? (13)

– T = 42.46 × 60 × 60 = 152856 s (2m)

– 2

T or V

RT

2, V = 17334 m s–1 (2m)

– 152856

22 T

= 4.11 × 10–5 rad s–1 or

R

V = 4.11 × 10–5 rad s–1 (3m)


What is the radius of Io if the acceleration due to gravity on the surface of Io is 1.796 m s–2?

– g

GMR

R

GMg ,

2 (2m)

– 796.1

)10931.8)(1067.6( 2211

R (2m)

– 1.821 × 106 m = 1821 km (2m)



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(a) (cont’d.)

Ganymede is the largest moon of Jupiter, and Io completes 4 orbits for each orbit of Ganymede. Calculate the radius of Ganymede’s orbit. (9)

– GM

RT

32 4

– 3

I

3G

2I

2G

R

R

T

T or

– TG = 4TI = 4(152856) = 611424 s (3m)

– 38

3G

2I

2I

)10217.4(

)4(

R

T

T = 1.90 ×1027 kg or

J

3I2

I4

GM

RT

→ MJ = 1.90 ×1027 kg (3m)

– 3 38G )10217.4(16 R = 1.063 × 109 m or

– J

3G2

G4

GM

RT

→ RG = 1.063 × 109 m (3m)


(mass of Io = 8.931 × 1022 kg)


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(b) What is meant by the critical angle of a substance? (3 × 2m) (6)

– angle of incidence when – angle of refraction is 90° – for light going from more dense to less dense medium

A diver at a depth under water in a lake observes a circle of light on the smooth surface of the lake. This circle of light is called Snell’s window.

Explain, using a ray diagram, why light is not observed at all points on the surface of the lake. (7)

Ray diagram showing

– light rays striking surface of water from above being refracted towards position of diver (2m)

– rays incident near 90° at edge of Snell’s window being refracted towards diver (2m)

– state that it is not possible for light to reach position of diver from outside Snell’s window (3m)

The area of the circle of light observed by the diver is 38.49 m2. (15)

Calculate

(i) the critical angle for water

– C

nsin

1 , sin C = 33.1

1 (3m)

– C = 48.75° (3m)


(ii) the depth of the diver under water

– 38.49 = πr2 and r = 3.5 m (3m)

– h

rC tan (3m)

– h = 75.48tan

5.3 = 3.07 m (3m)

(refractive index of water = 1.33)


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(c) What is a thermometric property? (2 × 3m) (9)

– a property that changes – measurably with temperature

What is the thermometric property of a thermistor? (3m)

– electrical resistance

Explain how a thermistor could be converted to a temperature gauge. (4 × 3m) (19)

– thermistor (attached to ohmmeter) and mercury-in-glass thermometer in some apparatus whose temperature is capable of being varied and measured, e.g. beaker of water

– water heated in steps of 10 °C / water at 0 °C and 100 °C (2 fixed points) – measure temperature and resistance for each reading – plot graph of resistance vs. temperature / ohmmeter scale modified to temperature

This temperature gauge was then used to measure the temperature of a heating plate in a sealed vacuum chamber.

State two advantages of using a thermometer based on a thermistor instead of a mercury-in-glass thermometer or a thermocouple for this application. (4m + 3m)

– difficult for mercury-in-glass thermometer to access or be read in sealed chamber / ohmmeter connected to thermistor can be external to chamber

– only one junction required for thermistor / no need to maintain second junction at fixed temperature


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(d) Answer either part (i) or part (ii).

(i) State two differences between leptons and hadrons. (9)

– leptons are fundamental particles (3m); hadrons are not fundamental particles but are made up of particles called quarks (3m)

– leptons don’t experience and hadrons do experience the strong nuclear force (3m)

Hadrons can be split into two subcategories.

Give the quark structure of a particle in each subcategory. (12)

Baryons

Any 1: (3m) – proton → uud // – neutron → udd // – lambda → udc // – sigma → dds

Mesons

Any 1: (3m)

– pion → du //

– kaon → sd

Give the quark structure of the anti-particle of one of these particles.

Any 1: (3m)

Baryons

Any 1:

– anti-proton → duu //

– anti-neutron → ddu //

– anti-lambda → cdu //

– anti-sigma → sdd

Mesons

Any 1:

– anti-pion → du //

– anti-kaon → sd


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(d) (i) (cont’d.)

Deduce the charge of this anti-particle.

Any 1: (3m)

Baryons

Any 1:

– anti-proton → –1

3

1,

3

2,

3

2//

– anti-neutron → 0

3

1,

3

1,

3

2 //

– anti-lambda → –1

3

2,

3

1,

3

2//

– anti-sigma → +1

3

1,

3

1,

3

1

Mesons

Any 1:

– anti-pion → –1

3

1,

3

2//

– anti-kaon → 0

3

1,

3

1

The electron was the first of the leptons discovered in 1897.

Name another lepton and explain why it took much longer to discover than the electron. (7)

Lepton

Any 1: (3m) – muon // – tau // – neutrino

Why it took so long to discover

Any 1: (4m) – muon and tau particles have a very short lifetime // – neutrino particles have very small mass, no charge


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(d) (ii) What is meant by rectification in an electrical circuit? (2 × 3m) (6)

– the conversion of alternating current (a.c.) – to direct current (d.c.)

Sketch graphs showing the output voltage for (i) a half-wave rectifier and (ii) a bridge rectifier, when the input is connected to an a.c. source. (10)

(i) a half-wave rectifier

Graph showing

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

–

(ii) a bridge rectifier

Graph showing

– truncated sine wave (2m) – no gaps in wave pattern (either smoothed or unsmoothed wave) (3m)

–


–V

0 V

+V

Time

Rectifier output

–V

0 V

+V

Time

Rectifier output

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(d) (ii) (cont’d.)

Draw the circuit diagram for a bridge rectifier. (12)

Circuit diagram showing (3 × 3m)

– a.c. supply connected correctly to – diamond shape of 4 diodes – correct orientation of diodes

–

Why would you include a capacitor in this circuit? (3m)

– (a capacitor) smooths the a.c. output


–

+

load

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