Welsh Joint Education Committee s3

MS3 WELSH JOINT EDUCATION COMMITTEE £3.00 CYD-BWYLLGOR ADDYSG CYMRU

General Certificate of Education Tystysgrif Addysg Gyffredinol Advanced Subsidiary/Advanced Uwch Gyfrannol/Uwch

MARKING SCHEMES

SUMMER 2004

PHYSICS

INTRODUCTION

The marking schemes which follow were the ones used by the WJEC for the SUMMER 2004 examination in the GCE PHYSICS papers . They were finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conferences were held shortly after the papers were taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conferences was to ensure that the marking schemes were interpreted and applied in the same way by all examiners.

It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conference, teachers may have different views on certain matters of detail or interpretation.

The WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes. PH1 Mark Scheme – June 2004

Notes: This marking scheme, whilst reasonably complete does not give all the answers, which were credited by the examiners. It is hoped that the scheme is self-explanatory, though it will need to be read alongside the question paper. The following clarifications may be of use: Statements in brackets [ ] are exemplification, alternatives or statements which, whilst desirable in an answer were not required on this occasion for full marks. The numbers in parentheses ( ) are the marks, usually 1, for each response. e.c.f. stands for error carried forward, and indicates that the results of a previous (incorrect) calculation will be treated as correct for the current section. i.e. the mistake will only be penalised once. The expression “or by implication” indicates that the mark is credited when subsequent credit-worthy working demonstrates that this idea/equation has been used. Incorrect or absent units only attract a penalty when ((unit)) appears, otherwise they are there in the mark scheme for completeness. No penalties for excessive significant figures are applied in this paper.

Marks Question Answers / Explanatory notes available 1 (a) (i) Attempt to use Pythagoras Thm. (1)

42 4 2 FR 9  0  10  9  0  10 (1) 5 3 FR 1  27  10 N (1)

(ii) Horizontal vector to right drawn on the dotted line. 1

5 1 iii) 1ּּ 27  10 N)

(b) (i) F ma (1)(or by impl.) 1 27  105 (e.c.f.) d  0  25m s-2 ((unit))(1) 2 5 0  105

v2 u 2  2 as [or by impl.] (1) Or alternative (ii) method involving 36 64  2  0  25s calculation of t 28 etc. s  (1)[correct sign and re-arrangement] 0  5 56  0m (1) 3

1 Marks Question Answers / Explanatory notes available 2 (a) Deceleration [or equiv.] 1

total distance  travelled s mean speed  or (b) (i) total time  taken t 1 4 2 (ii) mean speed  (1)  0  4m s-1 (1) 2 7 2

(c) (i) [Instantaneous] speed (or speed at a given time) 1

(ii) Tangent drawn (at 6ּּ 0 s) (1) Gradient correctly found  speed = 0ּּ 42 m s-1. (1) 2 d) L marked at any point where gradient < at t = 6ּּ 0 s. 1)

(e) No, with explanation (1). Train would travel at infinite speed 1 [or equiv.] (1) [10]

3 (a) Either (i) the [vector] sum (1)[or equiv.] of all the forces [acting on the body] = 0 (1) Or (ii) as above for torques/moments (2) 2

(b) (i) Vertical downwards vector drawn at G. (1) W = mg = 49 N (10 2

A (ii) 49 N e.c.f. B C

3 Diagram (1). T  /5 (1)= 49 N (1 for equation) (T= 81ּּ 7 N (1 [or alternative method e.g. by moments] 4

(c) Horizontal arrow from CB 1

Horizontal component of tension in AC in direction BC. 1 [10]

2 4 (a) (i) To provide an appreciable/measurable extension (accuracy implication) 1

(ii) Will be affected to the same degree by temperature variations. 1

(iii)  Readings repeated (during removal of load)  Diameter measured at a number of places along wire (and mean found) Any 2   Diameter found by taking readings at right (1) angles. 2  Parallax argument.

(iv) Load at which the wire will not return to the original length [on removal of load] (or equiv.) 1 b) (i) A = 8ּּ 0  10-8 m2 [accept 0ּּ 08 mm2] 1)

stress Fl  (ii) E  or  strain Ae  l 20 1  8 gradient (1)  A 5 103 8  10  8  10 9  0  10 Pa [90 GPa] (1) 4

5 (a) Alternating clear and opaque strips(1) /gaps/lines/scratches/ holes [in a barrier] which are equally spaced (1) 2

(b) (i) n d sin  (1) sin  [or equiv. in words] (1) Red has longest  (1) 3

(ii) X correctly labelled on the 2nd Y line 1

(iii) (Diffraction) angle of 3rd order blue is less than (diffraction) nd angle of 2 order red (or equiv.) (1). Because λblue<λred. (1) 2

(iv) nblue  3(1) 3(e.c.f.)  3  0  106 sin 26  7 0   449nm (1) 2

3 6 (a) (i) Observer sees blows instantaneously / sound waves take some time [or longer than light] to reach observer. (1) Sound heard corresponds to previous blow seen. (1) 2

660 (ii) Speed   330m s-1 (ans.) [N.B. other answers poss.] 2 1

(iii) Co-incidence broken (or equiv.) 1

(iv) Yes, at 1320 m [or other quantitative qualification] (1) Sound heard now corresponds to 2nd previous blow (1). [N.B. Other answers possible.] 2

(b) (i) ½ cycle in 0ּּ 1 s (1) or 1½ cycles (1) So T = 0ּּ 2 s (1) or 0ּּ 067 s (1) {or other correct ans.} 4 1 (ii) f  (or by impl.) (1). T F calculated correctly using T from (b)(i) 2

(iii) I. nodes 1 II. Position correctly identified 1

(iv) From diagram 3(1)  2  0m.    0  67 m(1) [no e.c.f.] -1 c f (or by impl.)(1)  c  3  3m s (1)[or other correct ans.] 4

(c) (i) Q labelled correctly 1

(ii) R labelled correctly 1

4 7 (a) (i) Polarised – (direction of) oscillations (1) (or displacement) (of electric field) confined to one plane (1). Unpolarised – (direction ..) … in random (many/all) planes (1) 3

(ii) The observer sees alternate light (1) and dark (1). There are two positions of light (or dark) in each 1800.(1) 3

(i) c (b) f  (1)[or by impl.]  3 0  108 f  4  0  1014 Hz(1) 2 750 109 n c (ii) I. 1 2 (1)[or by impl] (1) n2 c 1 1 00 c 2 (1) c  2  0  108 m s -1 1 50 3  0  108 2 2 2 0  108  (1)  500 nm (1) II. 4 0  1014 2

(iii) Light reflected from a ‘pit’ has travelled further than light reflected from a ‘bump’. (1) Path difference (further distance) = 250 nm (1) = λ/2 (1) Therefore in antiphase / completely out of phase… (1) 4

(c) (i) Acceleration is the [rate of] change of velocity (1). Velocity is constantly changing (1) because the direction is changing (1). 3

(ii) Towards the centre [of the disk]. 1

5 PH2 Mark Scheme – June 2004

Notes: This marking scheme, whilst reasonably complete does not give all the answers, which were credited by the examiners. It is hoped that the scheme is self-explanatory, though it will need to be read alongside the question paper. The following clarifications may be of use: Statements in brackets [ ] are exemplification, alternatives, indications of acceptable range of numerical answers (with ), or statements which, whilst desirable in an answer were not required on this occasion for full marks. The numbers in parentheses ( ) are the marks, usually 1, for each response. e.c.f. stands for error carried forward, and indicates that the results of a previous (incorrect) calculation will be treated as correct for the current section. i.e. the mistake will only be penalised once. The expression “or by implication” indicates that the mark is credited when subsequent credit-worthy working demonstrates that this idea/equation has been used. Incorrect or absent units only attract a penalty when ((unit)) appears, otherwise they are there in the mark scheme for completeness. No penalties for excessive significant figures are applied in this paper.

Question a) (i) 2ּּ 4 A 1) 1 ii) 19ּּ 2 A 1) iii) 0ּּ 625 Ω 1)

(b) Rseries = 4ּּ 0 Ω so y = 0ּּ 30 A (1) x  64(1) [e.c.f. from (a)(ii)] y 2

l (c) (i) R  (1)[or equiv.] A or by impl. l A  (1)[algebra] R 3  74  109 m 2 (1) 3

3 74  109 (e.c.f.) (ii) t  (1) 1 5  10-3 2  5  106 m (1) 2

6 2 (a)  Suitable means of heating e.g. water bath with coil immersed (1).  Thermometer shown in water bath or equiv. (1)  Basic procedure: heat, take readings, heat more etc. (1)  At least one precaution to promote accurate temperature readings. (1)  Resistance measurement: Ammeter-voltmeter: Direct reading: Connection of meters (1) Meter & range (1) V orConnected across coil (no R from R  (1) separate battery) (1) 6 I

R R (b)    0 (1) or by impl. R0

R0 11  0  0  5  (1) 3° -1 -1 4  5  5  10 (1) C or K  (1) [10]

3 (a) For a [metallic] conductor at constant temperature (1) the current is [directly] proportional to the applied p.d. (1) 2

8 4 R100  (1) or I  0  12 A (b) (i) SB 12 R 70  (1) SB 2 ii) 3ּּ 6 V [no e.c.f.] 1) (c) (i) [The current in AS] increases (1), because the total resistance of the circuit is decreased [or equiv.] (1) 2

(ii) [The p.d. between A and S] increases because VAS= I.RAS [in which RAS is constant] (1) 1

(iii) [The p.d. between S and B] decreases (1), because VAS increases and VAS + VSB = 12 V (e.c.f.) (1) 2

7 4 (a) (i) The [total] energy transferred by the cell (1) [or converted to electrical potential energy] per unit charge [or coulomb] [passing through the cell] (1).

(ii) The p.d. across the tell terminals [or the load].

(iii) The p.d. across the internal resistance. b) (i) I. 1ּּ 62 V) 4 II. recognisable intermediate step; e.g. ‘lost volts’ = (V (1)r = 0ּּ 60 Ω (1 27 0ּּ 1 1 (ii) P EI (1)[or by impl. is 1  62 V has been used] 0  73W (1) [e.c.f. on I ] 2 [iii) P = 0ּּ 61 W [e.c.f. on I) 2

5 6 0  1022 (a) n 3(1)  (1) 0 01m3 1  8  1029 m -3 ((unit))(1) 3

2 (b) A  2  0  104 m 2 (or by impl.) v1  4  104 m s -1 (1) [e.c.f. on n and diam/rad slip] 2

(c) (i) [Mean] energy of vibration of ions [accept atoms] increases (1) [accept: random energy of free electrons increases] Collisions between [free] electrons and ions more frequent (1), so drift velocity decreases. (1) 3

(ii) I decreases, as I = nAve (1) [with nAe constant] V  R increases since R  (1) [with V constant]. I  2 [10]

8 6 (a) When e-m radiation [accept light, u-v, photons] [of high enough frequency] falls on a surface [accept ‘metal’](1) electrons are emitted (1) [from the surface]. 2

I (b) (i) I. Number per second (1) or by impl.  5  6  1011 s -1 (1) e 2 II. Light (1) [or radiation / u-v / photons; accept KE of electrons] provides energy for electrons to flow [through wires](1). 2 1 .ii) I. 1ּּ 2 V) 1 .II. 1ּּ 9  10-19 J e.c.f

(c) (i) Graph: Energy scale (1); points (1); line (1) 3 (ii) I. the [minimum] energy needed to release an electron 1 [from the surface].

II. the minimum frequency of light needed to release an 1 electron from the surface.

1 mv2  hf  (1) or equiv. 2 max III. 1 When f f , mv2  0(1) 2 02 max

14 [IV. graph extrapolated to f = 0 or f0 = [4ּּ 55  10 Hz So  = 3ּּ 0  10-19 J (1) 2 .(1)

(iii) Correct method attempted [i.e. graph slope] (1) or subst. Of  value and a pair of (f, KEmax) values in equation. H = 6ּּ 6  0ּּ 210-34 J (2); [0ּּ 7 (1)] – accepted only if arising out of clear working. 3 [20]

9 7 (a) (i) Loss of electron[s] from an atom [or gain] 1

(ii) A proton. 1 iii) I. 13ּּ 6 eV or 2ּּ 2  10-18 J ((unit)) 1)

II. (any two from) gamma, X-ray, u-v, [visible] 2

III. photon energy too low. 1

(b) (i) Graph with vertical spikes [at least one pair] (1), a hump (1) and a non-zero threshold wavelength (1) 3

(ii) I. Threshold wavelength indicated. 1

II. Tube (accelerating) voltage [accept [kinetic] energy of the incident electrons]. 1

(c) (i) E 2  1 eV (1) [or by impl.] 2  1  1  6  1019 J(1) hc c E Use of  or   and f  (1) E f h 7  5  9  10 m (1) [powers of 10 lose 1 mark only] 4

(ii) I. Dark lines (1) crossing continuous spectrum /band of colours(1) 2

(II. 5ּּ 9  10-7 m (e.c.f.) (1 [photons of] this wavelength absorbed (1), promoting electrons from ground state into first excited state (1) [or 3 photons re-emitted in random directions]. [20]

10 PH3 Mark Scheme – May 2004

Notes: This marking scheme, whilst reasonably complete does not give all the responses, which were credited by the examiners. It is hoped that the scheme is self-explanatory, though it will need to be read alongside the question paper. The following clarifications may be of use:

Statements in brackets [ ] are exemplification, alternatives, indications of acceptable range of numerical answers (with ), or statements which, whilst desirable in an answer were not required on this occasion for full marks.

The numbers in parentheses ( ) are the marks, usually 1, for each response.

By the nature of a practical examination, the data are the candidates own and every attempt is made not to penalise candidates unduly for poor results, especially in the sections involving their analysis. The various sections of the questions are independent. e.c.f. stands for error carried forward, and indicates that the results of previous (incorrect) calculation or poor measurement will be treated as correct for the current section, i.e. the mistake will only be penalised once. This does not extend to errors of principle, for example inappropriately drawing a best-fit line through the origin and subsequently stating that the intercept is zero.

In this mark scheme, credited responses are sometimes grouped and bulleted. Where this occurs, it is understood that all the bulleted points must be present for full credit to be given, and each omission will have been penalised by one mark. This method of marking is only used in the PH3 papers.

Marks Experiment Available 1. (a) (i) Point in middle of ruler (1) 2 Diagonals /axes of symmetry drawn to show exact centre (ii) Centre of gravity [accept centre of mass] 1 (b) (i) Correct distance (to nearest mm) with unit 1 (ii) 02  10 = 2 N [accept 196 N] 1 (iii) Correct calculation [(i)  (ii)] - no unit penalty 1 (iv) Principle of Moments stated [allow lack of reference to a particular point] (1) 2 Weight correctly calculated(1) (c) (i) All readings taken [table complete] (1) [Allow 4 sets of readings, if the ruler is too heavy] All readings to 1 d.p. [rule resolution] (1) 4 Moment and weight calculated correctly (1) Moment and weight quoted to 2 or 3 s.f. (1)

(ii) Smallest distance from pivot/ 55 cm pivot (1) 2 1 mm error gives largest % error [or equiv.] (1) (iii) Correct calculation (1) 2 Accurate vlaue to 2 or 3 s.f. (1) [no unit penalty]

(d) (i) Correct calculation [no unit penalty] (1) 2

11 (ii) 5% of average calculated (1) Compared with (d)(i) and reliability correctly reported (1) 2 [or equivalent method using the uncertainty formula]

(iii) Take more repeat readings or use a larger range. 1 Total mark [20]

2. (a) Weight, [normal] reaction and friction all marked (2) One force omitted or incorrect 1 2 [accept: W / gravity / weight / mg (not g), R, air resistance / drag] (b) (i) time correct [within 1 s of centre’s value] (1) 2 Calculation correct [ignore s.f.s and unit] (1) (ii) Table: All units correct (1) s.f. consistency in data (1) 4 Calculation of sin  (1) Calculation of acceleration (1)

(c) Graph: Axis labelled with units [e.c.f. from table - penalise if units on sin ] (1) Non-awkward scales occupying at least ½ paper (1) 5 All points plotted correctly [within ½ minor square] (1) Graph passes through (0,0) (1) Line of best fit drawn (1)

(d) Good size triangle on graph (1) 3 Gradient calculated correctly (1) with correct unit (1) (e) (i) gradient = g or 10 [accept 9.8 – no unit penalty] (1) 2 intercept = 0 (1) (ii) Rotation of marble / friction / air resistance [N.B. It is recognised that the most significant contribution to the lower value of acceleration is that the sphere gains rotational, as well as translational, 1 kinetic energy. As this concept is not covered in the specification, it is anticipated that most candidates will refer to friction] (f)

Any reasonable answer e.g. light gates to measure time / a release mechanism. 1 Not: more readings. Accept: “reduce friction” only if by specified means.

Total mark [20]

12 3. (a) Symbols correct (1) 2 Circuit correct (1) (b) Correct range and units 1 (c) (i) Table: Tidy table (1) All columns labelled (1) Units correct for all headings (1) Good range and sample size (1) 6 At least 5 sets of readings (1) At least two d.p. for both I and V (1)

(ii) Power calculated correctly (1) 2 Resistance calculated correctly (1) (d) (i) Graph: Axes drawn, labelled and correctly oriented (1) Reasonable (non-awkward) scale occupying  ½ paper (1) Graph originates at (0.0) (1) 6 Points plotted correctly (–1 for each incorrect) (2) Reasonable smooth curve drawn (1)

(ii) [Marked together] & Correct values from graph [no unit penalty] (1) 1 (iii) Indicating values on graph(1) 1

(iv) A2 or units on ordinate  units on abscissa / W –1 [e.c.f.] 1 Total Mark [20]

13 PH4 Mark Scheme – June 2004

Notes: This marking scheme, whilst reasonably complete does not give all the answers, which were credited by the examiners. It is hoped that the scheme is self-explanatory, though it will need to be read alongside the question paper. The following clarifications may be of use: Statements in brackets [ ] are exemplification, alternatives, indications of acceptable range of numerical answers (with ), or statements which, whilst desirable in an answer were not required on this occasion for full marks. The numbers in parentheses ( ) are the marks, usually 1, for each response. e.c.f. stands for error carried forward, and indicates that the results of a previous (incorrect) calculation will be treated as correct for the current section. i.e. the mistake will only be penalised once. The expression “or by implication” indicates that the mark is credited when subsequent credit-worthy working demonstrates that this idea/equation has been used. Incorrect or absent units only attract a penalty when ((unit)) appears, otherwise they are there in the mark scheme for completeness. No penalties for excessive significant figures are applied in this paper.

Question Answers / Explanatory notes Marks available 1 (a) (i) Molecules hit container wall (1) [randomly]. Their changes in momentum mean they [experience forces from the wall and so] exert [equal and opposite] forces on the wall. 2

(ii) As temperature rise so does [mean] speed [or mean K.E.] of molecules (1), so hits on walls are more frequent [or equiv.] (1) [or exert greater force]. 2

1 2 (b) (i) p 3  c [or by impl.] (1) 0 80   kg m-3 [or by impl.] (1) 0 10 3 c2  470 m s -1 (1)

(ii) pV nRT (1) [or by impl.] 0 8 n   25 mol 0 032 3 3 T  290K ((unit)) (1)  e.c.f. on slips of 10  [10]

14 2 (a) A collision in which no kinetic energy is lost [or gained]. 1

[b) (i) 0ּּ 10 kg  50 ms-1 (1); 0ּּ 30kg  10 ms-1(1) [indiv. momenta) Total momentum = 2ּּ 0 kg ms-1 (1) 3

(ii) Use of ½ mv2 (1)  140 J (1) 2

0 30v  0  10  31  2  0(1) [e.c.f. from (b)(i)] (iii) I. v 17 m s-1 (1) 2 II. Final K.E. = 91 J (1) e.c.f. [Final < Initial] inelastic (1) [e.c.f. only if reasoning correct] 2 [Or: correct argument based on relative velocity of separation c.f. approach] [10]

3 (a) (i) Angular velocity [of a body about circle centre]. 1 [Accept: angular speed or frequency]

(ii) Use of v r or equivalent. 1

v2 F m (1)  3600 N (1) (b) (i) r 2

-2 3 (ii) 2300 (1) kg ms [or N] (1) South (1)

(iii) Mean (1) force (1) [on car going from A to B] [Or equivalent, e.g. the constant force (1) which, acting over the 7ּּ 9 s, would give the car the same change in momentum(1)] 2 [10]

15 4 (a) (i) F-1 [or V C-1] 1

(ii) Energy [stored in capacitor for a charge Q] Or Work [done charging capacitor to charge Q] 1

(b) (i)  RC (1)  50s (1) 2 Q Q et /  (1) [or equiv or by impl.] (ii) I. 0

t ln Q / Q0  (1) or equiv. e.g. t   ln 2  35s (1) 3 2 1 Q II. U  2 (1) [or equiv. or by impl.] C 1 2 Fraction of original  4 (1)

III. Internal [or thermal / random] energy in R or heat given 1 out by R [or both!] [10]

5 (a) (i) Increase in [accept ‘change in’] internal energy. 1

(ii) Work done by the system. 1

(b) (i) Heat is being lost from the block (1) at [almost] the same rate that work is being done on it (1). 2

(ii) I. Tangent drawn correctly (1). Slope calculation method correct (1) [even if inaccurate or wrong tangent drawn].  0ּּ 1C s-1. (1) 2 1 1ּּ

II. U  mc  (1)   ans. consistent with previous ans. (1) 3

III. Heat loss negligible [because block is at same temperature as the surroundings]. 1

16 6 (a) Acceleration  displacement [accept distance] from a fixed point (1) and directed towards it (1). 2

(b) (i) I. T 2 m k (1)[or equiv. or by impl.] 4 2m k  (1)[re-arranging] T 2 12 0s But T  (1) [or by impl.] 20 k  22 N m-1 ((unit))(1) 4

2 A v A (1) or [or by impl.] II. max T -1 0  31m s (1) e.c.f. on T  2

4 2 A a A 2 (1) or  or by impl. III. max T 2 3  3m s-2 (1) e.c.f. on  or T  2

(ii) I. (new length  natural length) [or equiv.] 1 II. mg k  extension (1) [or equiv. or by impl.] -2 g = 9ּּ 5 m s (1) [e.c.f. on k] 2

III. Obeys Hooke’s Law [accept; not reached elastic limit] 1 or massless /weightless.

Any specific thing which would alter the surface area (c) (i) appreciably [or surrounding with liquid (1) Surface area increased /fluid more viscous/dense [or equiv.] (1) Air resistance increased / resistance to motion increased / systems’s energy reduced. (1) 3

Graph drawn which: starts at maximum (1), has half cycles of (ii) equal durations (1) and amplitude decays ~ exponentially (1) 2

17 2 7 (a) (i) V I. Use of or equiv. even if wrong V is used. R V V  0 (1) or mean power = ½ max. power. r. m . s . 2 Mean power = 12 W. (1) 3

12W 1 s (1)  0  24 W (1) e.c.f. on 12 W II. 50   2

(ii) 100 (1) because the peak power occurs at both extremes of p.d. waveform (1) [or direction of current irreverent] 2 Vor IR or R (b) R At 90 to R (1)

Labelling of VC and VR Vor IZ or I R2 X 2 vectors (1)  C [whichever way round] 2 2 orZ or R X C Formula established (1) [with resultant vector shown on diagram or reference to Pythagoras]. 3 1 X (1)  15  9k  (c) (i) C C

Z18  8k  (1) [e.c.f. on X C ] 10 0V I (1)  5  3  104 A  Z   V IR (1)  5  3V OUT 4

XC approaches 0 (1), so VC approaches 0 [or equiv.] (1) (ii) 2 Curve drawn through (0,0) (1), through (1000, 5ּּ 3) having (iii) correct asymptote(1). 2 At high frequencies output ~ input. At low frequencies output (iv) p.d. «input p.d. 1

18 PH5 – Mark Scheme – June 2004

Notes: This marking scheme, whilst reasonably complete does not give all the answers, which were credited by the examiners. It is hoped that the scheme is self-explanatory, though it will need to be read alongside the question paper. The following clarifications may be of use: Statements in brackets [ ] are exemplification, alternatives, indications of acceptable range of numerical answers (with ), or statements which, whilst desirable in an answer were not required on this occasion for full marks. The numbers in parentheses ( ) are the marks, usually 1, for each response. e.c.f. stands for error carried forward, and indicates that the results of a previous (incorrect) calculation will be treated as correct for the current section. i.e. the mistake will only be penalised once. The expression “or by implication” indicates that the mark is credited when subsequent credit-worthy working demonstrates that this idea/equation has been used. Incorrect or absent units only attract a penalty when ((unit)) appears, otherwise they are there in the mark scheme for completeness. Significant figures may be examined in questions requiring data analysis.

Marks Question Answers / Explanatory notes available 1 (a) A A0 exp  t  (1)

A0  ln    t A A0 n   Or equiv. using  2 A 1 A0    ln  (1) 2 t A  1 (b) No – activity large [c.f. background]

ln 2 (c) T1 (1)  21800s (1) 2 2 

(d) A0  N 0 (1) 12 N0 1  74  10 (1) 2

(e) (i) 12 27 M Nm (1)  1  74  10  100  1  7  10  2  9  1013 kg (1) 2

(ii) Highly dangerous – e.g. small amount is very radioactive [or other sensible comment]. 1 [10]

19 Marks Question Answers / Explanatory notes available a) (i) Attempting LHS – RHS (1). 0ּּ 0052 u (1) 2) 2

(ii) E  mc2(1)  7  8  10 13 J(1) 2 or E  m  934(1)  4  9MeV(1)

1 2 13 (iii) 2 mv 7  8  10 (1) 2E v (1)  1  5  107 m s -1  2 m

mv2 mv (b) Bqv(1) (1).  r  (1) re-arranging  0  21m(1) 4 r Bq [10]

3 (a) (i) 1 1 N (1)   3  1026 (1) m 2 0136  1  7  1027 2 26  or: No. of moles 1000 2(1). N  500  N A  3  10 (1) 

(ii) mass defect = 0ּּ 0257 u (1) E = mc2 = 3ּּ 9  10-12 J [or E = 0ּּ 0257  934 = 24 MeV] (1) 2

(iii) Energy released = energy per reaction  number of reactions (1) 3 1026 =3 9  1012   5  7  10 14 J(1) 2[required] 2 [.alt  3ּּ 6 1033 eV or equiv]

 No radioactive products (b) Any 2  (1)  Lots of energy  Plenty of Deuterium/hydrogen available 2

(c) No force on motion along B. [or equiv.] (1) Circular motion around field lines with reason or force perpendicular to motion for perpendicular component (1). 2

20 4 (a) dI dI E  L or  or  changing flux (1) dt dt dI Large [or equiv.]/ flux changes rapidly (1) 2 dt

dI (b) (i) 0 [or implied] (1); E  0(1) 2 dt

(ii) dI 0 2  (1) dt 3 106 dI 3 E 55  103  (1)  3700V(1) dt

V 3700 (c) 4 E(1); d 6  7  3  10 m (1) d 5 10 3 Separation must be less than [or equal to] this (1) [10]

5 1 Q1 Q 2 EP  (1) 40 r e2 (1) r  5  3  1011 m(1) 3 40 EP

1 Q Q 1 2 8 2 F 2 (1)  8  2  10 N(1) [allow e.c.f. on (2e) ] 2 40 r

M M F G 1 2 (1)  3  6  1047 N (1) No, much smaller (1) 3 r 2

Both attractive (1), because of + and – and gravity is always attractive [allow: masses positive] (1) 2

21 6 (a) (i) Path shown along axis of accelerator. 1

(ii) By E-field [or voltage] between the tubes. 1

(iii) Particles move faster [in successive tubes](1). Larger tubes needed to keep frequency constant [or to stay in synch.](1). 2

(b) (i) Path shown spiralling outwards (1) anticlockwise (1). 2

(ii) Between the Dees [accept anywhere in the B-field] 1

(iii) Constant frequency [or impl.] (1), for constant B field(1). [Accept 2 sensible independent comments] 2

(c) (i) Path shown circling (1) clockwise(1). 2

(ii) X marked between tubes 1

(iii) Increasing the voltage [or decreasing the gap or E-field] 1

(iv) B-field [intensity] increased 1

1 2 (d) (i) 2 mv eV (1) 1 31 2  19 2 9  1  10v  1  6  10  300000[subst.] (1) v 3  2  108 m s -1 (1) 3

(ii) Mass [of protons] (far) greater (1), so speed (much) smaller (1) and there is less of a problem (1). accept calculated speed (7ּּ 6106 m s-1) + comment] 3] [20]

22 7 (a) (i) h  (1)  2  9  1011 m (1) p 2

hc (ii) c fand E  hf or   (1) E E 10000  1  6  1019 (1)  1  24  1010 m (1) 3

1 mv2 10keV [energy] (1) [or equiv. or by impl.] (iii) 2 p mv (1)[or equiv. or by impl.]  1  23  1011 m (1) 3

(b) [e.c.f.] all are appropriate (1), since λ is similar / less than / of 2 the order of interatomic spacing (1)

(c) (i) E 57  4keV (1)  57  4  103  1  6  10 19 J(1) hc 3  (1)  2  2  1011 m  E  

n = fringe number (order) (ii) 4 correct  (2) d = (slit) spacing 2 or 3 correct  (1) λ = wavelength θ = diffraction angle 2

2 6 (iii) tan  or sin  or   (1) 50 n 1(1) 2  17  1011 d   4  2  1010 m (1) 3 sin 0 052

(iv)  Good long-range order or low defect concentration Any 2  Crystal not cubic crystal orthogonal independent  Vertical and horizontal separation of marks. planes different.  Vertical plane separation > horizontal (2) 2

23 PH6 – Mark Scheme – June 2004

Notes: This marking scheme, whilst reasonably complete does not give all the answers, which were credited by the examiners. It is hoped that the scheme is self-explanatory, though it will need to be read alongside the question paper. The following clarifications may be of use: Statements in brackets [ ] are exemplification, alternatives, indications of acceptable range of numerical answers (with ), or statements which, whilst desirable in an answer were not required on this occasion for full marks. The numbers in parentheses ( ) are the marks, usually 1, for each response. e.c.f. stands for error carried forward, and indicates that the results of a previous (incorrect) calculation will be treated as correct for the current section. i.e. the mistake will only be penalised once. The expression “or by implication” indicates that the mark is credited when subsequent credit-worthy working demonstrates that this idea/equation has been used. Incorrect or absent units only attract a penalty when ((unit)) appears, otherwise they are there in the mark scheme for completeness. Significant figures may be examined in questions requiring data analysis.

Marks Question Answers / Explanatory notes available a) (i) 5ּּ 0 [ 0ּּ 1] m s-1 1) 1

(ii) 0 [zero] 1 b) 5ּּ 0 m s-2 [ignore sign] 1)

(c) gradient changes suddenly / stops instantaneously (1) Very large / infinite acceleration (1) or equiv. 2

(d) Curved modification to graph 1

(e) Any two correct points. (1 + 1) Correct plotting (1) with smooth curve showing increasing positive gradient. (1) 4

2 (a) (i) 86 1

(ii) 136 1

(b) Radon can be breathed in (or equiv – accept the converse). Accept: radon emits higher energy α-particles, or that radon is more unstable. 1

1 (c) (i) mv22  4  10 6  1  6  10 19 (1) 2 m4u(1)  v  1  07  107 m s -1 (1) 3

24 V Ed (1)  2  1  106  0  06  12  6  10 4 V (1) 2 (ii) F qE E a(1)  (1)  2 e (1)  1  1014 m s -2 3 m m m (iii) d 0 18 8 2 t (1) 7  1  7  10 s (1) (iv) v 1 07  10

1 2 x ut  2 at (1) or equiv using two equations (v) x 1  45 cm. (1) 2 [15]

3 (a) 1 2 EK  2 mv(1)so K.E. more sensitive to v than m . (1) [Or K.E.  m (1) and  v2 (1)] 2

(b) accelerates electrons (or equiv.) 1

2 (c) Definition requires: per second, per m and energy of X-rays (any 2) 2

100mA (d) Number (1)  6  25  1017 (1) 1 6  10-19 1

(e) x 2cm (  1cm)(1) [or implied] v2 u 2  2 as [or equiv.] (1) v2 a  6  1017 m s -2 (1) 2x 3

(f) Stop build-up of charge (or equivalent) 1

Energy levels with electron dropping (2) (g) [or electron drops (1), from one energy level to another (1)] 2

(h) P = IV [so PI, if V is constant.] 1

(i) P = IV so expect P  V (1). But as the voltage increases, the tube is more efficient so a smaller fraction of the energy is dissipated as heat in the metal. (1) 2

(j) Diagrams show that an inner electron is ejected (1) by a collision (1). An outer electron drops (1) emitting an X-ray (1) 4

25 4 -1 1  -1 (a) LHS = [f] = s (1) RHS =   [L] = Ωs [or VA s](1) LC  s 1 [C] = RHS   equation is OK. (1) 4  s2

(b) (i) 5 values for table calculated (2) [3or4  (1)] to 2 or 3 sf (1) 3

1 1 - (ii) Unit = or F 2 1 F

(c) Sensible relating of equation to y mx(  c )(1) 1

(d) Axes (1), points (e.c.f.) (2), line (1) 4

(e) Straight line (1), through the origin (1). All points but one close to the line (1)(e.c.f.) So good agreement except for one point. (1) 4

y 1 m (1)  27  2 Hz F2 [ignore units!](1) 2 (f) (i) x

1 (ii)  27(e.c.f.)(1). 2 L 2 1  5 L   [rearranging](1)  3  5[  0  2]  10 H((unit))(1) 3 2  27  Graph drawing: Tidy curve through points  2. Peak above (g) (i) highest data point (1). 3 Needs more data (1) around peak (1). (ii) 2 .(Resonant frequency ~ 3ּּ 8 kHz (1 (iii) 1 From other graph, 140(1) [or equiv.] C C = 51 [6] μF (1) 3

N.B. The PH6 Investigatory Task component is marked according to the scheme in the subject specification. For comments about interpretation, please refer to the examiner’s report.

26 27 Welsh Joint Education Committee 245 Western Avenue Cardiff. CF5 2YX Tel. No. 029 2026 5000 Fax. 029 2057 5994 E-mail: [email protected]. website: www.wjec.co.uk/exams.html