NATIONAL SENIOR CERTIFICATE GRADE 12

Transcription

1 NATIONAL SENI CERTIFICATE GRADE 1 PHYSICAL SCIENCES: PHYSICS (P1) NOVEMBER 010 MEMANDUM MARKS: 150 This memorandum consists o 3 pages. NOTE: Marking rule 1.5 was changed according to decisions taken at the memorandum discussion, November 010.

2 Physical Sciences/P1 DBE/November 010 Learning Outcomes and Assessment Standards LO 1 LO LO 3 AS 1..1: Deine, discuss and explain prescribed scientiic knowledge. AS 1.1.1: Design, plan and conduct a scientiic inquiry to collect data systematically with regard to accuracy, reliability and the need to control variables. AS 1.3.1: Research, discuss, compare and evaluate scientiic and indigenous knowledge systems and knowledge claims by indicating the correlation among them, and explain the acceptance o dierent claims. AS 1.1.: Seek patterns and trends, represent them in dierent orms, explain the trends, use scientiic reasoning to draw and evaluate conclusions, and ormulate generalisations. AS 1.1.3: Select and use appropriate problem-solving strategies to solve (unseen) problems. AS 1.. Express and explain prescribed scientiic principles, theories, models and laws by indicating the relationship between dierent acts and concepts in own words. AS 1..3: Apply scientiic knowledge in everyday lie contexts. AS 1.3.: Research case studies and present ethical and moral arguments rom dierent perspectives to indicate the impact (pros and cons) o dierent scientiic and technological applications. AS 1.3.3: Evaluate the impact o scientiic and technological research and indicate the contribution to the management, utilisation and development o resources to ensure sustainability continentally and globally. AS 1.1.4: Communicate and deend scientiic arguments with clarity and precision.

3 Physical Sciences/P1 3 DBE/November 010 GENERAL GUIDELINES 1. CALCULATIONS Marks will be awarded or: correct ormula, correct substitution, correct answer with unit. No marks will be awarded i an incorrect or inappropriate ormula is used, even though there may be relevant symbols and applicable substitutions. When an error is made during substitution into a correct ormula, a mark will be awarded or the correct ormula and or the correct substitutions, but no urther marks will be given. I no ormula is given, but all substitutions are correct, a candidate will oreit one mark. When no ormula is given, marks will be oreited or zero substitutions not shown. Other substitutions and a correct answer will be credited. No penalisation i zero substitutions are omitted in calculations where correct ormula / principle is given correctly. Mathematical manipulations and change o subject o appropriate ormulae carry no marks, but i a candidate starts o with the correct ormula and then changes the subject o the ormula incorrectly, marks will be awarded or the ormula and the correct substitutions. The mark or the incorrect numerical answer is oreited. Marks are only awarded or a ormula i a calculation has been attempted. i.e. substitutions have been made or a numerical answer given. Marks can only be allocated or substitutions when values are substituted into ormulae and not when listed beore a calculation starts. All calculations, when not speciied in the question, must be done to two decimal places.. UNITS.1. Candidates will only be penalised once or the repeated use o an incorrect unit within a question or sub-question. Units are only required in the inal answer to a calculation.

4 Physical Sciences/P1 4 DBE/November Marks are only awarded or an answer, and not or a unit per se. Candidates will thereore oreit the mark allocated or the answer in each o the ollowing situations: - Correct answer + wrong unit - Wrong answer + correct unit - Correct answer + no unit.4 SI units must be used except in certain cases, e.g. V m -1 instead o N C -1, and cm s -1 or km h -1 instead o m s -1 where the question warrants this. 3. GENERAL I one answer or calculation is required, but two given by the candidate, only the irst one will be marked, irrespective o which one is correct. I two answers are required, only the irst two will be marked, etc. For marking purposes, alternative symbols (s,u,t, etc.) will also be accepted Separate compound units with a multiplication dot, not a ull stop, or example, m s -1. For marking purposes m.s -1 and m/s will also be accepted. 4. POSITIVE MARKING Positive marking regarding calculations will be ollowed in the ollowing cases: Sub-question to sub-question: When a certain variable is calculated in one sub-question (e.g. 3.1) and needs to be substituted in another (3. or 3.3), e.g. i the answer or 3.1 is incorrect and is substituted correctly in 3. or 3.3, ull marks are to be awarded or the subsequent sub-questions. A multi-step question in a sub-question: I the candidate has to calculate, or example, current in the irst step and gets it wrong due to a substitution error, the mark or the substitution and the inal answer will be oreited. I a inal answer to a calculation is correct, ull marks will not automatically be awarded. Markers will always ensure that the correct/appropriate ormula is used and that workings, including substitutions, are correct. Questions where a series o calculations have to be made (e.g. a circuit diagram question) do not necessarily always have to ollow the same order. FULL MARKS will be awarded provided it is a valid solution to the problem. However, any calculation that will not bring the candidate closer to the answer than the original data, will not count any marks.

5 Physical Sciences/P1 5 DBE/November I one answer or calculation is required, but two given by the candidate, only the irst one will be marked, irrespective o which one is correct. I two answers are required, only the irst two will be marked, etc. Normally an incorrect answer cannot be correctly motivated i based on a conceptual mistake. I the candidate is thereore required to motivate in question 3. the answer given to question 3.1, and 3.1 is incorrect, no marks can be awarded or question 3.. However, i the answer or e.g is based on a calculation, the motivation or the incorrect answer in 3. could be considered.

6 Physical Sciences/P1 6 DBE/November 010 SECTION A QUESTION Elastic [1..1] (1) 1. Huygens (principle) [1..1] (1) 1.3 ohm / Ω [1..1] (1) 1.4 (Split-ring) commutator [1..1] (1) 1.5 Work unction [1..1] (1) [5] QUESTION.1 A [1..3] (). C [1.1.] ().3 D [1..3] ().4 D [1..] ().5 D [1..1] ().6 C [1..] ().7 C [1..] ().8 B [1.1.] ().9 C [1..1] ().10 A [1..] () [0] TOTAL SECTION A: 5

7 Physical Sciences/P1 7 DBE/November 010 SECTION B QUESTION seconds / 3 s [1.1.] (1) 3. v = u + at Accept the equations: s = ut + 1 at v + u s = t v = u + as OPTION 1 Area between graph and time axis y = (area o triangle)/ ½ bh = ½ (3)(9,4) = 44,1 m Maximum height above ground: ,1 = 144,1 m OPTION v + v i v + v i Δy = Δt Δy = Δt 0 + 9,4 9,4 + 0 = 3 = 3 = 44,1 m (43,m) Maximum height above ground: ,1 = 144,1 m (143,m) OPTION 3 From edge o cli to max height (Upward positive) v = v i + aδy 0 = 9,4 + (-9,8)Δy Δy = 44,1 m Maximum height above ground: ,1 = 144,1 m From edge o cli to max height) (Downward positive) v = v i + aδy 0 = (-9,4) + (9,8)Δy Δy = - 44,1 m Maximum height above ground: ,1 = 144,1 m

8 Physical Sciences/P1 8 DBE/November 010 OPTION 4 From edge o cli to max height (Upward positive) 1 Δ y = v iδt + aδt = (9,4)(3) + ½(-9,8)(3) = 44,1 m Maximum height above ground: ,1 = 144,1 m (143, m) From edge o cli to max height) (Downward positive) 1 Δ y = v iδt + aδt = (-9,4)(3) + ½(9,8)(3) = - 44,1 m Maximum height above ground: ,1 = 144,1 m OPTION 5 From max height to edge o cli Downward positive v = v i + aδy (9,4) = 0 + (9,8)Δy Δy = 44,1 m Maximum height above ground: ,1 = 144,1 m OPTION 6 From max height to edge o cli Downward positive 1 Δ y = v iδt + aδt = (0)(3) + ½(9,8)(3) = 44,1 m Maximum height above ground: ,1 = 144,1 m OPTION 7 E mech (edge o cli) = E mech (max height) (mgh + ½ mv ) A = (mgh + ½ mv ) B any equation m(gh + ½ v ) A = m(gh + ½ v ) B (9,8)(100 )+ ½ (9,4) = (9,8)h + 0 h = 144,1 m OPTION 8 W net = ΔE k mghcosθ = ½ m(v v i ) any equation m(ghcosθ) = ½m(v - v i ) OPTION 9 E mech (edge o cli) = E mech (max height) (mgh + ½ mv ) A = (mgh + ½ mv ) B m(gh + ½ v ) A = m(gh + ½ v ) B any equation (9,8)hcos180 o = ½ (0 h = 44,1 m (43,m) - (9,4) ) 0 + ½ (9,4) = (9,8)h + 0 h = 44,1 m Maximum height above ground: ,1 = 144,1 m Maximum height above ground: ,1 = 144,1 m [1.1.] [1..3] (4)

9 Physical Sciences/P1 9 DBE/November Checklist: Criteria or graph Correct shape t = 3 s at maximum height t = 6 s at y = 0 m Graph starts at y = 0 m at t = 0 s Marks position (m) position (m) time (s) [1.1.] (4)

10 Physical Sciences/P1 10 DBE/November OPTION 1: Upward positive: v = v i + a t = 9,4 + (-9,8)(5,3) = -1,85 m s -1 downwards v = 1,85 m s -1 downwards OPTION 1 Δ y = v iδt + aδt = 9,4 (5,3) + ½ (-9,8)(5,3) = 19,73 m (or both ormulae) v = v i + aδy = 9,4 + (-9,8)(19,73) v = 1,85 m s -1 downwards POSITIVE MARKING FROM 3.1 OPTION 3 (Downward motion only) Downward positive: Time or downward motion: (5,3 3) =,3 s v = v i + a t = 0 + (9,8)(,3) = 1,85 m s -1 downwards Downward positive: v = v i + a t = -9,4 + (9,8)(5,3) = 1,85 m s -1 downwards Downward positive: 1 Δ y = v iδt + aδt = (-9,4)(5,3) + ½ (9,8)(5,3) = -19,73 m v = v i + aδy (or both ormulae) = (-9,4) + (9,8)(-19,73) v = 1,85 m s -1 downwards OPTION 4 Downward positive: Time or downward motion: (5,3 3) =,3 s 1 Δ y = v iδt + aδt = (0)(,3) + ½ (9,8)(,3) = 4,3671 m (or both ormulae) v = v i + aδy = (0) + (9,8)(4,3671) v = 1,85 m s -1 downwards [1..3] (5) 3.4. POSITIVE MARKING FROM QUESTION OPTION 1 Upward positive: t = (5,3 1) = 4,3 s v = v i + a t = 49 + (-9,8)(4,3) v = 7,55 m s -1 upwards v XY = v X - v Y (vector dierence) = -1,85-7,55 = -9,40 m s -1 downwards v XY = 9,40 m s -1 downwards v XY = v XG + v GY = -1,85 + (-7,55) = -9,40 m s -1 downwards v XY = 9,40 m s -1 downwards v XG = v XY + v YG -1,85 = v XY + (7,55) = -9,40 m s -1 downwards v XY = 9,40 m s -1 downwards

11 Physical Sciences/P1 11 DBE/November 010 Downward positive: t = (5,3 1) = 4,3 s v = v i + a t = (9,8)(4,3) = -7,55 m s -1 v = 7,55 m s -1 upwards OPTION Upward positive: t = (5,3 1) = 4,3 s 1 Δ y = v iδt + aδt = 49(4,3) + ½ (-9,8)(4,3) = 119,59 m (upwards) v = v i + aδy (or both equations) = (49) + (-9,8)(119,59) v = 7,55 m s -1 upwards v XY = v XG + v GY = 1,85 + (7,55) v XY = 9,40 m s -1 downwards v XY = v X - v Y (vector dierence) = 1,85 (- 7,55) = 9,40 m s -1 downwards v XG = v XY + v YG 1,85 = v XY + (-7,55) = 9,40 m s -1 downwards v XY = v XG + v GY = -1,85 + (-7,55) = -9,40 m s -1 downwards v XY = 9,40 m s -1 downwards v XY = v X - v Y (vector dierence) = -1,85-7,55 = -9,40 m s -1 downwards v XY = 9,40 m s -1 downwards v XG = v XY + v YG -1,85 = v XY + (7,55) = -9,40 m s -1 downwards v XY = 9,40 m s -1 downwards Downward positive: t = (5,3 1) = 4,3 s 1 Δ y = v iδt + aδt = (-49)(4,3) + ½ (9,8)(4,3) = -119,59 m (upwards) v = v i + aδy (or both equations) = (-49) + (9,8)(-119,59) v = 7,55 m s -1 upwards v XY = v XG + v GY = 1,85 + (7,55) = 9,40 m s -1 downwards v XY = v X - v Y (vector dierence) = 1,85 (-7,55) = 9,40 m s -1 downwards v XG = v XY + v YG 1,85 = v XY + (-7,55) = 9,40 m s -1 downwards [1.1.3] (5) [19]

12 Physical Sciences/P1 1 DBE/November 010 QUESTION 4.1 The sum o the kinetic and (gravitational) potential energy is conserved / constant / remains the same / does not change in an isolated / closed / system / no external work done / only conservative orces act on the system. The (total) mechanical energy is conserved/ constant in an isolated system. [1..1] () 4. OPTION 1 E mech = U + K or E p + E k = mgh + ½ mv (any ormulae) = (0,5)(9,8)(0,6) + ½ (0,5)(3) = 5,19 J (5,5 J) OPTION E p = mgh = (0,5)(9,8)(0,6) =,94 J (3 J) E k = ½ mv = ½ (0,5)(3) =,5 J E mech = E p + E k =,94 +,5 = 5,19 J 4.3 Accepted ormulae E mech(a) = E mech(b) / E mech(i) = E mech() / E mech(top) = E mech(bottom) (E p + E k ) A = (E p + E k ) B / (E p + E k ) bottom = (E p + E k ) top (E p + E k ) i = (E p + E k ) /(U + K) bottom = (U + K) top (U + K) i = (U + K) / (U + K) A = (U + K) B / mgh i + 1 mv = mgh + i 1 mv [1..3] (4) OPTION 1 (U + K) B = (U + K) C mgh B + ½ m v B = mgh C + ½ m v C 5,19 = 0 + ½ (0,5)v v = 4,56 m s -1 OPTION W net = ΔE k mgδycosθ = ½ m(v v i ) (0,5)(9,8)(0,6)(1)= ½ (0,5)(v 3 ) v = 4,56 m s -1 Σ p beore = Σ p ater (0,5)(4,56) + 0 = (0,5)v + (0,1)(3,5) v = 3,86 m s -1 (to the right) (3,88 m s -1 ) Other ormulae: p t beore = p t ater or m 1 v i1 + m v i = m 1 v 1 + m v or m 1 u 1 + m u = m 1 v 1 + m v [1.1.3] [1..3] (7) [13]

13 Physical Sciences/P1 13 DBE/November 010 QUESTION 5 N w F applied F horizontal / F x / F // F vertical / F y / F Accepted Labels Normal / Force o surace on crate / F N / 69 N / 75 N F g / orce o Earth on crate / weight / 94 N /300 N mg / gravitational orce F / orce o worker on crate / 50 N / F A F riction / 0 N / F / riction 43,30 N 5 N 5.1 N 30 o w F applied Accept: Force diagram N w F vertical F horizontal N 30 o F applied N F vertical w w F horizontal [1.1.] (4) 5. W = F x cos90 o = 0 They (normal orce and the gravitational orce) are perpendicular /at 90 o to the (direction o the) displacement / motion / Δxo the crate. The angle between the orce and displacement / motion / Δx is 90 o. The crate moves horizontally and the orces act vertically. [1..] ()

14 Physical Sciences/P1 14 DBE/November Accepted symbols or applied orce: F appl / F / F A Accepted symbols or rictional orce: / F / F riction Accepted symbols or gravitational orce: w / F g / F orce o Earth on crate / gravitational orce OPTION 1 W net = W appl + W For either ormula = F app Δx cos θ + Δx cos θ = (50)(6)(cos30 ) + (0)(6)(cos180 ) = 59,81 + (-10) W net = 139,81 J OPTION W applied = F app Δx cos θ = (50)(6)(cos30 ) = 59,81 J W = Δx cos θ = (0)(6)(cos180 ) = -10 J W net = W applied + W F app Δxcosθ + FΔxcosθ = 139,81 J OPTION 3 W net = W appl // + W For either ormula = F app// Δx cos θ + Δx cos θ = (50)(cos30 )(6)cos 0 + (0)(6)(cos180 ) = 59,81+ (-10) W net = 139,81 J OPTION 4 F net = F horizontal + = (50)(cos30 ) + (-0) = 3,30 N W net = F net x cos θ = (3,30)(6)(cos 0 ) = 139,81 J OPTION 5 F net = F horizontal + ma = (50)(cos30 ) + (-0) (30)a = (50)(cos30 º ) + (-0) a = 0, m s - v = v i + aδx = (0) + (0,78...)(6) v = 3,05... m s -1 W net = K = ½ m(v v i ) = ½ (30)(3, ) = 139,81 J one mark or all three ormulas [1.1.3] (4)

15 Physical Sciences/P1 15 DBE/November W net = K / W net = E k = ½ mv - ½ mv i 139,81 = ½ (30)v 0 v = 3,05 m s -1 I: W instead o W net max ( ) 3 No marks or any other method [1..3] (3) 5.5 Greater than The horizontal component (o the orce) / orce in direction o motion will now be greater / F net will now be greater. As θ decreases cos θ increases For θ smaller than 30, cos θ > cos 30. [1.3.] () [15] QUESTION Doppler eect [1..1] (1) 6. v ± v v + v = L L L = s / L s v ± v s v vl = 340 v L = 14,17 m s -1 (960) [1..3] (4) 6.3 Higher than [1..] (1) [6]

16 Physical Sciences/P1 16 DBE/November 010 QUESTION When two waves pass through the same region o space at the same time, resulting in the superposition o waves. [1..1] () 7. Constructive (intererence) The waves crossing each other are in phase. / Two troughs meet./ The path dierence is an integer number o λ. [1.1.] () 7.3 Dark band - - It lies on the line combining all the points where crests and troughs overlap resulting in destructive intererence. It lies on the (nodal) line where destructive intererence occurs. [1.1.] (3) [7]

17 Physical Sciences/P1 17 DBE/November 010 QUESTION The ability o a wave to bend / spread out (in wave ronts) as they pass through a (small) aperture / opening or around a (sharp) edge/ points /corners / barrier. () Angle o / (Degree o) diraction Position o minima α or β [1.1.1] (1) 8.. (Slit) width / a [1.1.1] (1) 8.3 (Slit) 1 Slit 1 represents the most diraction. - Diraction /Angle / sin θ / θ is inversely proportional to slit width. 1 1 sin θ α or θ α a a Larger angle at which irst minimum or slit 1 is obtained. Smaller angle at which irst minimum or slit is obtained. Actual calculations showing slit 1 is narrower than slit. [1.1.] [1..] () 8.4 OPTION 1 mλ sin θ = a 9 (1)( ) sin 5 = a a = 4,70 x 10-6 m (0, m / 4,7 μm) OPTION mλ sin θ = a 9 ()( ) sin 10 = a a = 4,7 x 10-6 m (0, m / 4,7 μm) OPTION 3 Allocated ull marks i calculation shown correctly in QUESTION 8.3. [1.1.3] (4) [10]

18 Physical Sciences/P1 18 DBE/November 010 QUESTION The ratio o the (amount o) charge (transerred) to the (resulting) potential dierence. [1..1] () 9. C = ε 0 A Κε or C = 0 A where Κ = 1 d d 3 x (8,85 10 )( 10 )(1,5 10 ) = 3 1,5 10 C = 1,77 x 10-1 F (1,77 pf) [1..3] (5) 9.3 I (A) Checklist Criteria or graph: Correct shape o graph showing decreasing current with time. Mark/ Punt t(s) [1.1.] () 9.4 C = V Q 1,77 x 10-1 = 1 Q Q =,1 x C [1..3] (3) 9.5 OPTION 1 Vq F = d 19 (1)(3, 10 ) = 3 1,5 10 =,56 x N OPTION V 1 E = = = 8 x 10 3 V m -1 d 3 1,5 10 F E = q 8 x 10 3 F = 19 3, 10 F =,56 x N [1.1.3] (5) [17]

19 Physical Sciences/P1 19 DBE/November 010 QUESTION The current in a conductor is directly proportional to the potential dierence across its ends at constant temperature. The ratio o potential dierence to current is constant at constant temperature [1..1] () R p 1 1 = + = R R Rp = 0,7 Ω 1,4 1,4 R p = R1R R + R 1 = 1,4 1,4 1,4 + 1,4 = 0,7 Ω [1..3] (3) 10.. OPTION 1: em = I(R + r) 1 = I(0,7 + 0, 1) I = 15 A R = I V 0,7 = 15 V V = 10,5 V OPTION 3 Voltage divides 0,7: 0,1 / 7:1 V headlight = 8 7 x 1 = 10,5 V OPTION : V 1 I = = = 15 A R 0, 8 V = IR = (15)(0,7) = 10,5 V em = I(R + r) 1 = V external + (15)(0,1) V external = 1 (15)(0,1) = 10,5 V V lost = Ir = (15)(0,1) = 1,5 V V external = 1 1,5 V = 10,5 V 15 I headlight = = 7,5 A V = IR = (7,5)(1,4) = 10,5 V [1..3] (4)

20 Physical Sciences/P1 0 DBE/November OPTION 1 P = R V 10,5 = 1,4 = 78,75 W OPTION I(light) = 7,5 A P = VI = (10,5)(7,5) = 78,75 W OPTION 3 I(light) = 7,5 A P = I R = (7,5) (1,4) = 78,75 W OPTIONS ACCEPTED ONLY BECAUSE BULBS ARE IDENTICAL: OPTION 4 P total = R V (10,5) = 0,7 = 157,5 W P 157,5 headlight = OPTION 5 P total = VI = (10,5)(15) = 157,5 W P 157,5 headlight = = 78,75 W OPTION 6 P total = I R = (15) (0,7) = 157,5 W P 157,5 headlight = = 78,75 W 10.3 Decreases - = 78,75 W (Eective/ total ) resistance decreases. (Total) current increases. Lost volts / V internal / Ir increases, thus potential dierence / V (across headlights) decreases. V P = decreases. R [1..3] (3) [1.1.3] (4) [16]

21 Physical Sciences/P1 1 DBE/November 010 QUESTION Vmax Vrms = = 311,13 = 0 V [1..3] (3) OPTION 1 P ave = V rms I rms 100 = (0)I rms I rms = 0,45 A Imax I rms = I max = 0,45 = 0,64 A OPTION P ave = 100 = V rms R (0) R R = 484 Ω R = V I max max 311, = I max I max = 0,64 A OPTION 3 P ave = V rms I rms V I = max max = V I max max,13 I 100 = I max = 0,64 A 311 max [1.1.3] (5)

22 Physical Sciences/P1 DBE/November potential dierence (V) Correct shape or at least one cycle and loops o equal height. time (s) potential dierence (V) Correct shape or at least one cycle and loops o equal height. time (s) [1.1.] () [10]

23 Physical Sciences/P1 3 DBE/November 010 QUESTION Any ONE: Damage to skin./causes (skin) cancer. Damage to eyes./increased occurrence o cataracts. Damage to crops resulting in ood shortages. [1.3.] (1) 1. Kills bacteria / germs / Sterilises/ sanitises / disinects equipment. [1.3.] (1) 1.3 OPTION 1 hc E = λ 34 8 (6,63 10 )(3 10 ) = = 9,95 x J OPTION c = λ 3 x 10 8 = (00 x 10-9 ) = 1, 5 x Hz E = h = (6,63 x )(1,5 x ) = 9,95 x J or both ormulae [1..3] (4) 1.4 OPTION 1 E = W o + E k For either ormula h = h o + ½ mv 9,95 x = 3,84 x ½ (9,11 x )v v = 1,16 x 10 6 m s -1 ( , ,94 m s -1 ) OPTION E = W o + E k E k = 9,95 x ,84 x = 6,11 x J Other symbols: E: h W o : h o K: E k; : ½ mv E k = ½ mv 6,11 x = ½ (9,11 x )v v = 1,16 x 10 6 m s -1 ( , ,94 m s -1 ) [1..3] (4) 1.5 Yes (Photons o) X rays have a higher requency / shorter wavelength / - energy (than ultraviolet radiation). UV light has lower requency than X-rays. [1..] () [1] TOTAL SECTION B: GRAND TOTAL: