2004 Physics. Higher. Finalised Marking Instructions

Transcription

1 004 Physics Higher Finalised Marking Instructions

2 Scottish Qualifications Authority Detailed Marking Instructions - Higher Physics. General Marking Instructions SQA published Physics General Marking Instructions in July 999. Please refer to this publication when interpreting the detailed marking instructions.. Recording of marks The following additional advice was given to markers regarding the recording of marks on candidate scripts. (a) (b) (c) (d) (e) (f) (g) The total mark awarded for each question should be recorded in the outer margin. The inner margin should be used to record the mark for each part of a question as indicated in the detailed marking instructions. The fine divisions of marks shown in the detailed marking scheme may be recorded within the body of the script beside the candidate s response. Where such marks are shown they must total to the mark in the inner margin. Numbers recorded on candidate scripts should always be the marks being awarded. Negative marks or marks to be subtracted should not be recorded on scripts. The number out of which a mark is scored should never be recorded as a denominator. (½ mark will always mean one half mark and never out of ) Where square ruled paper is enclosed inside answer books it should be clearly indicated that this item has been considered by the marker. The mark awarded should be transferred to the script booklet inner margin and marked G. The mark awarded for each question should be transferred to the grid on the back of the script. When the marker has completed marking the candidate s response to all questions, the marks for individual questions are added to give the total script mark. The total mark awarded for an individual question may include an odd half mark - ½. If there is an odd half mark in the total script mark, this is rounded up to the next whole number when transferred to the box on the front of the script. Page

3 004 Physics Higher Marking scheme Section A. C. B. D. E 3. D 3. E 4. C 4. C 5. B 5. D 6. B 6. D 7. A 7. A 8. A 8. D 9. E 9. B 0. E 0. C Page 3

4 004 Physics - Higher Sample Answer and Mark Allocation Notes. (a) speed is a scalar has magnitude/size only velocity is a vector has speed/size + direction vel needs direction () do not accept unit (b)(i) distance = 0 x x 5 = 7 km If give direction (0) (ii) 4 5 km at 39 W of N or 5 N of W (3) or 3 (± 0 4 km) and (± ) For partial marking look for Scale or lengths of lines proportional to distances Correct vector addition (arrows not required) 4 5 km (± 0 4 km) + 5 (± ) N of W 3 (± ) () 39 º (± ) W of N (½ + ½) 5 4 km no arrow 5º 4 5 km 5º (3) 4 5 km + 3 or 4 max () distance speed = time or 0 displacement vel = time no mention of time (0) Accept 5 + = 7 km wrong/no unit deduct no final statement deduct Allow 3º a = b + c bc cos A a = 5 + x5xcos0 a = 4 5 km (½ + ½) sin 0 sin θ = 4.5 θ = 5 N of W 3 () 39 W of N + Accept rounding to 5 km 3+ 9 (iii) Av. velocity = 7 v = (0) displacement time = 4.5 = 7 5 (km h - ) at 5 N of W consistent with b(ii) wrong unit (-½) d v = (on own) (0) t s v = (on own) t Page 4

5 (c) For Leeuvin v = t s 7 5 = 4.5 t t = 93 (h) Leeuvin -0 5 = 75 (h) s = vt = consistent with b(ii) h 56 min total time = = 8 (h) = 3 5 (km) Mir first h min + hence Mir is first Leeuvin -0 5 = 75 (h) If use 0 5 (h) for 5 minutes (WP) s v = = t from that point = 8 3 (kmh - ) Mir first Page 5

6 . (a) (i) a = v u t = = 5 (m s ) s = ut + at for both a and s equations s = 60 x 40 + ½ ( 5) x 40 s = 00 m + v = u + as for both a and s equations 0 = 60 + x ( 5) s s = 00 m + Do not need to show but must appear in nd equation If use u = 0 max Must be negative or WP u + v s = ( ) t s = x 60 x 40 s = 00 m + v u If s = ( ) t (WP) If s = ut + at = ( 5) 40 7 (a) (ii) [a = (b) v u t F = ma = = 5 m s ] F = 7 5 x 0 5 x ( ) 5 (signs must be consistent) F = ( ) 3 x 0 6 N + If add statement force E = Fd = 6 = 3 0 N () mv both equations needed F 00 = ( 60) 6 F = 3 0 N + P = IV Accept this formula 8 5 x 0 6 = 5 x 0 3 anywhere in the answer V V = 3 4 x 0 3 (V) Vp Accept this formula V rms = anywhere in answer 3 4 x 0 3 Vp = V p = 4 8 x 0 3 V + = 4808V + but V (max ½) (sig figs) Page 6 max for implied formula Similarly for v = u + as v, u mixed up max If F= ma shown acceleration consistent with (a) (i) unless 9 8 ms - Ft = mv mu F x 40 = x 0 5 x 60 F = N + If add weight (WP) max () E = IVt 8 5 x 0 6 = 5 x 0 3 x V x V = 3 4 x 0 3 (V) Q = It E E V = = Q It Ip I rms = (irrelevant) = 3 4 x 0 3 (A) Vp not V rms = 3+

7 3. (a) Tension = (weight) = mg = 5 x 0 4 x 9 8 Tension = 4 9 x 0 5 N + (b)(i) tension buoyancy (force) or upthrust g = 0ms - once per question W= mg = 4 9 x 0 5 N + but if continues on (WP) (0) All 3 labels + 3 correct direction () ( each incorrect or missing force) 7 symbols T, W, U (0) Weight pull of gravity force of gravity Gravitational force gravity, gravity force, uplift (0) (b) (ii) Upthrust = Weight Tension Upthrust = 4 9 x x 0 5 Upthrust = 4 x 0 5 (N) P = A F 5.4 x 0 P = 8 P = 3 0 x 0 4 Pa 5 F.5 x 0 If use P = = 4 = N A 8 4 = 3 0 N max consistent with (a) F P = P P P = F F A = A A x 0.5 x 0 = P = 3 0 x 0 4 Pa Each substitution Subtraction no unit or final line deduct (c) No change or difference in pressure same () difference in pressure depends on height/thickness/ shape difference and P = h ρ g or P depth as no thickness change (h -h ) same No explanation attempt (0) Pressure change between top and bottom is same () No change + explanation but not (WP) () + Do not accept size Page 7

8 4. (a) (i) lost volts = E V = = (V) lost volts = Ir = I x 0 I = 0 6 (A) V 7.8 R = = I 0. 6 R = 3 0 Ω + V R = V R = 7 8 R Remaining values in correct place ( or 0) R = 3 Ω If goes on eg 3 R = Ω deduct () max () (a) (ii) Current in internal resistor gives lost volts (so external voltage reduced) () External voltage or tpd = E Ir but as I increases and E and r constant then V decreases Voltage divider explanation eg voltage is divided across R and r so smaller reading on meter ( or 0)) No explanation (0) Lost volts on own (0) Voltage flowing/moving through (WP) (0) Voltage/energy lost across/in internal resistance (r) () Energy lost in the cell () Voltage lost in the cell () (b) (External) Resistors in parallel gives lower total resistance current increases so lost volts increases * reading on voltmeter decreases * look to see if this appears in answer before starting awarding marks If voltmeter reading increases/or same then (0) by calculation consistent with a(i) flowing voltage (-½) + Page 8

9 5. (a)(voltage across capacitor)/v 9 Shape () V units and 00 ms labels origin omitted deduct 0 00 (time)/ms V 0 t () V (0) 0 t/ms (0) (b) (i) V R = IR = 0 x 0 3 x 400 V R = 8 (V) If V = 8 V max V c = 8 V c = 4 V + (b) (ii) E = ½ CV E = ½ x 00 x 0 6 x 4 E = 8 x 0 4 J + Q = CV = 00 x 0 6 x 4 = 4 x 0 4 (C) E = ½QV for both Q and E equations E = 0 5 x 4 x 0 4 x 4 E = 8 x 0 4 J + consistent with b (i) No square or 6 shown formula only List of formulae (0) (if no selection made) (c) resistor value less than 400 Ω () Use larger voltage supply but do not charge above V () Remove resistance () Remove resistor (0) Remove resistor and close the gap () smaller lower smaller lower resistor (0) resistance () Use larger supply only (0) Page 9

10 (d) charging current on for shorter time () (so smaller charge required to reach V) (so value of C) less/smaller than 00 µf () * look for this first smaller area under graph () (so less charge stored for same V) so smaller value of C () Smaller capacitor accept as bad form If discharge (WP) (0) If give 00 µf (0) No explanation (0) If value of C less and justification not (WP) () Page 0

11 Rf 6. (a) V o = (V V) R V o = R f V R V o = V o = 0 4 V + 3 ( ) V o = V gain (b) (Temperature rises, R thermistor decreases so) voltage across thermistor decreases (voltage across variable resistor/v ) increases (V -V ) becomes more positive or increases Bridge becomes more out of balance (V o [ = (V -V ) x gain] increases) transistor switches on relay switches on the alarm + mention of voltage flowing (WP) stop marking (c) V o = gain x V or R f (V V ) 0 7 = x (V 7 50) V = 7 56 (V) () From graph, temperature = 36 C + ± 0 05 o C R 36ºC on own () + Page

12 sinθ sinθair 6 7 (a) (i) n = or watch for sinθ sinθliquid = n = 4 (0) sin 8 sin 45 n = sin 45 n = 40 () (a) (ii) n x sin θ liquid = sin θ air (But sin θ liquid = constant and n blue > n red sin θ air is now greater hence θ air greater (than 8 ) Since 45 constant then n α sin θ air then sin θ air larger so θ air larger than 8 must have attempt at explanation before first awarded by calculation using n > 4 () Alternative Larger RI gives greater change in direction/bending so θ air > 8 θ air > 8 + explanation (not (WP) eg λ is smaller than λ in air + If use diffraction (0) (b) air degree sign missing (-½) Diagram with no calculation (0) () liquid Calculation + no diagram (max ) sin θ c = n sin θ c =.44 sin θair 44 = 0 sin 45 θ c = 44 ( ) () (43 (-½) arith) sin θ air > 45 > 44 so total internal reflection θ air = error/not possible () 3+ hence TIR Page

13 8 (a) (i) Radiation/photons of energy equal to difference between energy levels triggers an electron (in its excited state) to drop from higher to lower energy level with emission of an identical photon () photon of the same value/energy radiation Diagram (look for same 4 points) labelled electron in higher level incident photon labelled with energy difference level electron dropping to lower level photon emitted same energy difference or in phase with original atoms/particles falling (0) stop at (WP) 8 (a) (ii) Photons (are reflected by mirrors) light/radiation causes/triggers/to induce more electrons to drop from higher to lower energy level () Amplification is produced by a series of stimulated emissions () More energy is given out than absorbed () Content statement Photons producing more photons () Photons give more chance of stimulated emission () (b) d sin θ = nλ * might be there by implication 37 d sin = * x 633 x 0 9 d = 99 x 0 6 (m) no. of lines per m = d = 5 0 x 0 5 (lines/m) () = 5 03 x 0 5 (lines/m) also possible d 5 (allow ) d sin θ = nλ sinθ nλ = d sin37 / = x 633 x 0 9 d = 5 0 x 0 5 () d If use 37 max 3+ Page 3

14 (c) (fringe width less) so θ (or sin θ) less so d sin θ less 3 but d and n are constant 4 therefore λ less, 3, 4 (), 3, 4 (),, 4, 4 () 3, 4 () 4 must be correct before any marks awarded, but must have attempt at explanation Alternative path difference = nλ path difference less n is constant λ is less by calculation d sin θ = nλ since d, n values same sin θ smaller so λ smaller use diffraction (0) λ longer/larger/same (0) + Page 4

15 9 (a) p-type ( ) n-type a.c. supply (0) 6 (+) (-) () or (0) or or + - If diode symbol drawn (0) (b) Electrons and holes (re)combine (at junction) energy released as photons photons given out light photons combine join together, combine, falls into hole Cannot get second without first Creating energy (0) for second forming electron/hole pairs (0) Electrons and holes meet (0) (c) (i) E = hf 3 68 x 0 9 = 6 63 x 0 34 f f = 5 55 x 0 4 (Hz) v = fλ for both E and v equations 3 x 0 8 = 5 55 x 0 4 λ λ = 5 40 x 0 7 m + E (= hf) = λ hc x0 9 = λ λ = 5 40 x 0 7 m nm (54nm) x 0 7 m Accept 5 40 x 0 7 m x 0 7 m Too many figures (-½) (c) (ii) E = QV 3 68 x 0 9 = 6 x 0 9 V V = 3 V + + Page 5

16 30 (a) (i) 9 = number of protons (in nucleus) () (ii) 35 = protons + neutrons (in nucleus) total number of nucleons in nucleus. () 9p + 43 n mass no (0) atomic no (0) 9 electrons (0) 6 (b) neutrons released are absorbed/captured by/collide with another Uranium nucleus producing another fission/more splitting () (more neutrons released to continue/repeat process) chain reaction on its own (0) ( or 0) not reactions if use atom (0) fission correct spelling (c) mass before 35 U mass after x Ce 3 4 x n 675 x Zr x n x x 0 7 (kg) x 0 7 (kg) mass loss = 0 37 x 0 7 (kg) E = mc Independent mark E = 3 7 x0 8 x (3 x 0 8 ) E = 3 35 x 0 J + Watch for single neutron only on RHS mass then x 0 7 kg 3 7 x 0 8 (kg) Truncating of masses or mass loss WP but E = mc available No square or 9 x 0 6 formula only 3 [END OF MARKING INSTRUCTIONS] Page 6