May 2018 Physics Standard Level Paper 2 - IB Documents

M18/4/PHYSI/SP2/ENG/TZ2/XX/MMarkschemeMay 2018PhysicsStandard levelPaper 28 pages

–2–This markscheme is the property of the InternationalBaccalaureate and must not be reproduced or distributed to anyother person without the authorization of the IB Global Centre,Cardiff.M18/4/PHYSI/SP2/ENG/TZ2/XX/M

Notes1.aitowards the centre «of the circle» / horizontally to theright Do not accept towards the centre of the bowl1.aiidownward vertical arrow of any length Judge the length of the vertical arrow by eye. The constructionlines are not required. A label is not requiredarrow of correct length Total1Neg:21.aiiiALTERNATIVE 1F N cos θ mg N sinθ dividing/substituting to get result eg:ALTERNATIVE 23right angle triangle drawn with F, N and W/mglabelled angle correctly labelled and arrows on forces incorrect directions tan θ correct use of trigonometry leading to the requiredrelationship F O mg AFmgtan θ(continued )

–4–M18/4/PHYSI/SP2/ENG/TZ2/XX/M(Question 1 continued)1.bAward [4] for a bald correct answermgv2 m tanθrAward [3] for an answer of 13.9/14«ms-1». MP2 omittedr R cos θ v 4gR cos2 θgR cos θ9.81 8.0 cos 22// sinθta nθta n 22v 13.4 / 13 «m s 1» 1.1.cdthere is no force to balance the weight/N is horizontal so no / it is not possible Must see correct justification to awardMP2 v « 2gR » 12.5 « ms 1 » speed before collisionAllow 12.5 from incorrect use ofkinematics equations2Award [3] for a bald correct answerAward [0] for mg (8 ) 2 mgh leading toh 4 m if done in one step.«from conservation of momentum» common speed after collision isspeed « v c 1initial2Allow ECF from MP112.5 6.25 m s 1 » 2vc26.252 h « » 2.0 « m » 2g 2 9.81Allow ECF from MP23

–5–Question2.aiAnswersa gas in which there are no intermolecular forcesM18/4/PHYSI/SP2/ENG/TZ2/XX/MNotesTotalAccept atoms/particles.ORa gas that obeys the ideal gas law/all gas laws at all pressures, volumes andtemperatures1ORmolecules have zero PE/only KE 2.aii2.aiiipV 5.3 105 2.1 10 4N « » 2.6 1022 23kT1.38 10 310«For one atom U 3kT» 1.38 10–23 310 / 6.4 10–21 «J» 2231Allow ECF from (a)(ii)Award [2] for a bald correct answer2 « 2.6 10U2.bi2.bii223 1.38 10 23 310 » 170 « J » 2p2 « 5.3 105 3Allow use of U pV22.1 10 4» 1.6 105 « Pa » 46.8 101«volume has increased and» average velocity/KE remains unchanged The idea of average must be included«so» molecules collide with the walls less frequently/longer time betweencollisions with the walls Decrease in number of collisions is notsufficient for MP2. Time must beincluded.«hence» rate of change of momentum at wall has decreased «and so pressure has decreased»Accept atoms/particles.2 max

–6–3.3.aaiiithe incident wave «from the speaker» and the reflectedwave «from the closed end»superpose/combine/interfere Allow superimpose/add upDo not allow meet/interact1Horizontal arrow from X to the right MP2 is dependent on MP1Ignore length of arrow13.aiiiP at a node 3.aiv4 0.30wavelengthis λ « » 0.40 « m » 3340f « » 850 «Hz » 0.403.biM18/4/PHYSI/SP2/ENG/TZ2/XX/M1Award [2] for a bald correct answer2Allow ECF from MP1sinθC1 3401500Award [2] for a bald correct answerθC 13 « » Answer must be to 2/3 significant figures to award MP2Award [2] for a bald answer of 13.12Allow 0.23 radians3.biicorrect orientation Do not penalize the lengths of A and B in the watergreater separation Do not penalize a wavefront for C if it is consistent with A and BMP1 must be awarded for MP2 to be awardedeg:2CBA airwater

–7–Question4.Answersthe work done per unit charge aM18/4/PHYSI/SP2/ENG/TZ2/XX/MNotesTotalAward [1] for “energy per unit charge provided by thecell”/“power per unit current”Award [1] for “potential difference across the terminals of thecell when no current is flowing”2Do not accept “potential difference across terminals of cell”in moving charge from one terminal of a cell to the other /all the way round the circuit 4.bithe resistance is proportional to length / see 0.35 AND1«.00» 2so it equals 0.35 80 « 28 Ω»4.biicurrent leaving 12 V cell is12 0.15 « A »80Award [2] for a bald correct answerOR E12 28 80E « 0.15 28 » 4.2 « V » 2Allow a 1sf answer of 4 if it comes from a calculation.Do not allow a bald answer of 4 «V»Allow ECF from incorrect current

wersNotesTotalAverage height 127 «m» mgh gh 9.81 127» 1.2 103 J kg 1 Specific energy « mUnit is essentialAllow g 10 gives 1.3 10 3 J kg-12Allow ECF from 110 m( 1.1 10 3 J kg 1 ) or 144 m( 1.4 10 3 J kg 1 )5.aiimass per second leaving dam is1.2 105 103 « 2.0 106 kg s 1 » 60rate of decrease of GPE is 2.0 106 9.81 127 2.49 109 «W » / 2.49«GW » 5.a5.biiiefficiency is «Do not award ECF for the use of 110 mor 144 mAllow 2.4 GW if rounded value usedfrom (a)(i) or 2.6 GW if g 10 is used31.8 » 0.72 / 72% 2.5water is pumped back up at times when the demand for/price of electricity is low 1

–9–6.aM18/4/PHYSI/SP2/ENG/TZ2/XX/M«most of» the mass of the atom is confined within a very smallvolume/nucleus 2 max«all» the positive charge is confined within a very small volume/nucleus electrons orbit the nucleus «in circular orbits» 6.bithe energy needed to separate the nucleons of a nucleusORAllow neutrons AND protons for nucleonsDon’t allow constituent parts1energy released when a nucleus is formed from its nucleons 6.biiQ 106 8.550 106 8.521 3.07 « MeV » 1« Q 3 MeV »6.ciline with arrow as shown labelled anti-neutrino/ν Correct direction of the “arrow” is essentialThe line drawn must be “upwards” from thevertex in the time direction i.e. above thehorizontaleg:16.ciiV W– 1

This markscheme is the property of the International Baccalaureate and must not be reproduced or distributed to any other person wi