Markscheme May 2017 Physics Higher level Paper 2

Transcription

1 M17/4/PHYSI/HP/ENG/TZ/XX/M Markscheme May 017 Physics Higher level Paper 3 pages

2 M17/4/PHYSI/HP/ENG/TZ/XX/M This markscheme is the property of the International Baccalaureate and must not be reproduced or distributed to any other person without the authorization of the IB Global Centre, Cardiff.

3 3 M17/4/PHYSI/HP/ENG/TZ/XX/M 1. a correct use of kinematic equation/equations Substitution(s) must be correct or 149 or 150 «m» b a 7 or.45 «m s» 11 Could be seen in part (a). Award [0] for solution that uses a 9.81 m s F «N»

4 4 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 1 continued) c ALTERNATIVE 1 «work done to launch glider» «04 kj» «work done by motor» «power input to motor» 80 or 80.4 or 81 k«w» 3 11 Award [ max] for an answer of 160 k«w». ALTERNATIVE use of average speed 13.5 m s 1 «useful power output» force average speed « » 3 power input 100 « » 80 or 80.4 or 81 k«w» 3 ALTERNATIVE 3 work required from motor KE work done against friction « ( )» 04 «kj» «energy input» power input work required from motor k«w»

5 5 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 1 continued) d direction of Award [1 max] if forces do not touch the dot, but are otherwise OK. motion lift force drag weight drag correctly labelled and in correct direction weight correctly labelled and in correct direction AND no other incorrect force shown

6 6 M17/4/PHYSI/HP/ENG/TZ/XX/M. a force/acceleration proportional to displacement «from equilibrium position» and directed towards equilibrium position/point and directed in opposite direction to the displacement from equilibrium position/point Do not award marks for stating the defining equation for SHM. Award [1 max] for a x with a and x defined. b i frequency of buoy movement time period of buoy or «Hz» 35 or 10.3 «s» or 10 «s» v x «or fx» 4.3 or T «m s 1»

7 7 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question continued) b ii peaks separated by gaps equal to width of each pulse «shape of peak roughly as shown» one cycle taking 10 s shown on graph output power Judge by eye. Do not accept cos or sin graph At least two peaks needed. Do not allow square waves or asymmetrical shapes. Allow ECF from (b)(i) value of period if calculated. 0 0 time / s c i PE of water is converted to KE of moving water/turbine to electrical energy «in generator/turbine/dynamo» idea of pumped storage, ie: pump water back during night/when energy cheap to buy/when energy not in demand/when there is a surplus of energy

8 8 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question continued) c ii 1 specific energy available «gh» «650 J kg» 1 mgh mv v gh v 1 73 «ms» Do not allow 7 as round from 7.8 c iii total energy «mgh » «J» Use of «J»gives 11 h. For MP the unit must be present. total energy (answer (c)(ii)) mv J time 11.1h or s c iv friction/resistive losses in pipe/fluid resistance/turbulence/turbine or generator «bearings» sound energy losses from turbine/water in pipe thermal energy/heat losses in wires/components Must see seat of friction to award the mark. Do not allow friction bald. max water requires kinetic energy to leave system so not all can be transferred

9 9 M17/4/PHYSI/HP/ENG/TZ/XX/M 3. a «CV» 6.3 «J» 1 b 1 e 100 t t ln «s» 3 c i Q c mt Allow ECF from 3(a) for energy transferred. 370 Jkg 1 1 K Correct answer only to include correct unit that matches answer power of ten. Allow use of g and kj in unit but must match numerical answer, eg: 0.37 J kg 1 K 1 receives [1]

10 10 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 3 continued) c ii ALTERNATIVE 1 some thermal energy will be transferred to surroundings/along connecting wires/to thermometer ü estimate «of specific heat capacity by student» will be larger «than accepted value» ü ALTERNATIVE not all energy transferred as capacitor did not fully discharge ü so estimate «of specific heat capacity by student» will be larger «than accepted value» ü max

11 11 M17/4/PHYSI/HP/ENG/TZ/XX/M 4. a «light» superposes/interferes pattern consists of «intensity» maxima and minima consisting of constructive and destructive «interference» 3 voltage peaks correspond to interference maxima b i 7 D If no unit assume m. 3 «s».1 10 «m» 3 d Correct answer only. b ii correct read-off from graph of 5 m s Allow ECF from (b)(i) 3 x v «ms» 3 t c i angular width of diffraction minimum 0.13 «0.06 rad» Award [1 max] for solution using 1. factor. 5.0 slit width «m» d 0.06 c ii «beyond the first diffraction minimum» average voltage is smaller OWTTE «voltage minimum» spacing is «approximately» same rate of variation of voltage is unchanged

12 1 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 4 continued) d «reflection at barrier» leads to two waves travelling in opposite directions ü mention of formation of standing wave ü maximum corresponds to antinode/maximum displacement «of air molecules» max complete cancellation at node position ü 5. a 4 These must be seen on the right-hand side of the equation. 4 He 86 Rn

13 13 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 5 continued) b i ALTERNATIVE 1 6 days is s activity after 6 days is A e A0 A A 0 or N0 states that index of e is so small that 15 1 A A 0 10 s A is 1 A 0 Award [1 max] if calculations/substitutions have numerical slips but would lead to correct deduction. eg: failure to convert 6 days to seconds but correct substitution into equation will give MP. Allow working in days, but for MP1 must see conversion of or half-life to day 1. ALTERNATIVE shows half-life of the order of 10 s or s converts this to year «1600 y» or days and states half-life much longer than experiment compared to experiment

14 14 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 5 continued) b ii ALTERNATIVE 1 Must see correct substitution or answer to + sf for MP3 use of A N0 0 conversion to number of molecules nn A initial activity «s» number emitted (643600) or.7 10 alpha particles 3 ALTERNATIVE use of N N e t 0 0 N0 n NA alpha particles emitted or alpha particles «number of atoms disintegrated N N N (1 e ) 0» 0

15 15 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 5 continued) c i alpha particles highly ionizing alpha particles have a low penetration power thin glass increases probability of alpha crossing glass decreases probability of alpha striking atom/nucleus/molecule c ii conversion of temperature to 91 K Do not allow reference to tunnelling. Allow ECF for from (b)(ii) p p or 0.84 «Pa»

16 16 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 5 continued) d electron/atom drops from high energy state/level to low state energy levels are discrete wavelength/frequency of photon is related to energy change or quotes E hf hc or E and is therefore also discrete e peer review guarantees the validity of the work means that readers have confidence in the validity of work 3 1

17 17 M17/4/PHYSI/HP/ENG/TZ/XX/M 6. a when an electric field is applied to any material «using a cell etc» it acts to accelerate any free electrons electrons are the charge carriers «in copper» 3 metals/copper have many free electrons whereas insulators have few/no free electrons/charge carriers b i area «9.310 m» radius m b ii Ppeak I peak 730 «A» V peak 1

18 18 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 6 continued) b iii resistance of cable identified as «64 3» Allow [3] for a solution where the resistance per unit metre is calculated using resistivity and answer to (a) (resistance per unit length of cable m ) a power seen in solution I plausible answer calculated using «plausible if in range 10 W m 1 to 150 W m 1 when quoted answers in (b)(ii) used» 31 «W m 1»ü Award [ max] if 64 used for resistance (answer3). V An approach from or VI using 150 kv is incorrect (award R [0]), however allow this approach if the pd across the cable has been calculated (pd dropped across cable is 1.47 kv). 3 c response to (b)(ii) 60«A» 1

19 19 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 6 continued) d wires/cable attract whenever current is in same direction Award [1 max] if response suggests that there is repulsion between cables at any stage in cycle. charge flow/current direction in both wires is always same «but reverses every half cycle» max force varies from 0 to maximum force is a maximum twice in each cycle e i higher voltage gives lower current «energy losses depend on current» hence thermal/heating/power losses reduced e ii laminated core Do not allow wires are laminated. 1

20 0 M17/4/PHYSI/HP/ENG/TZ/XX/M 7. a i 5 hc wavelength «m» 19 E If no unit assume m. 1 a ii «potential» energy is required to leave surface Do not allow reference to binding energy. Ignore statements of conservation of energy. all/most energy given to potential «so none left for kinetic energy» b 19 energy surplus J v «m s» max 31 c i «same intensity of radiation so same total energy delivered per square metre per second» light has higher photon energy so fewer photons incident per second Award [1 max] if surplus of J used (answer: m s 1 ) Reason is required 1 c ii 1:1 correspondence between photon and electron so fewer electrons per second current smaller Allow ECF from (c)(i) Allow ECF from MP to MP3. 3

21 1 M17/4/PHYSI/HP/ENG/TZ/XX/M 8. a potential is defined to be zero at infinity ü so a positive amount of work needs to be supplied for a mass to reach infinity ü b i V S GM so r VS «GM» constant because G and M are constants 1 r b ii 0 1 GM «J m kg» GPE at Earth orbit «J» Award [1 max] unless answer is to sf. Ignore addition of Sun radius to radius of Earth orbit.

22 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 8 continued) b iii ALTERNATIVE 1 Various approaches possible. GMm work leading to statement that kinetic energy, AND kinetic energy evaluated r 33 to be.7 10 «J» energy «PE KE answer to (b)(ii).7 10».7 10 «J» ALTERNATIVE 1 statement that kinetic energy is gravitational potential energy in orbit so energy answer to (b)(ii) «J» b iv «KE will initially decrease so» total energy decreases «KE will initially decrease so» total energy becomes more negative Earth moves closer to Sun max new orbit with greater speed «but lower total energy» changes ellipticity of orbit

23 3 M17/4/PHYSI/HP/ENG/TZ/XX/M (Question 8 continued) c centripetal force is required and is provided by gravitational force between Earth and Sun Award [1 max] for statement that there is a centripetal force of gravity without further qualification.