Markscheme May 2017 Physics Standard level Paper 3

Transcription

1 M7/4/PHYSI/SP/ENG/TZ/XX/M Markscheme May 07 Physics Standard level Paper 7 pages

2 M7/4/PHYSI/SP/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 M7/4/PHYSI/SP/ENG/TZ/XX/M Section A a in order to keep the temperature constant in order to allow the system to reach thermal equilibrium with the surroundings/owtte Accept answers in terms of pressure or volume changes only if clearly related to reaching thermal equilibrium max with the surroundings. b c d recognizes b as gradient calculates b in range Pa m to V H thus ideal gas law gives p H so graph should be «a straight line through origin,» as observed ba n = OR correct substitution of one point from the graph RT n = = Award [ max] if POT error in b. Allow any correct SI unit, eg kg s -. Answer must be to or SF. Allow ECF from (b).

4 4 M7/4/PHYSI/SP/ENG/TZ/XX/M e very large means very small volumes / very high pressures H at very small volumes the ideal gas does not apply OR at very small volumes some of the assumptions of the kinetic theory of gases do not hold

5 5 M7/4/PHYSI/SP/ENG/TZ/XX/M a 4π.60 g = = L T g = g( + ) = « =» L T OR.0 % For the first marking point answer must be given to at least dp. Accept calculations based on gmax = gmin = gmax gmin = hence g = (9.8 ± 0.) «ms» OR = g 0. «ms» b T T 0 =.0 θ max = Accept answer from interval 0 to 4.

6 6 M7/4/PHYSI/SP/ENG/TZ/XX/M Section B Option A Relativity a a set of coordinate axes and clocks used to measure the position «in space/time of an object at a particular time» OR a coordinate system to measure x,y,z,and t / OWTTE b i magnetic only there is a current but no «net» charge «in the wire» b ii electric only P is stationary so experiences no magnetic force relativistic contraction will increase the density of protons in the wire

7 7 M7/4/PHYSI/SP/ENG/TZ/XX/M 4 a ΔtP / observer sitting in the train b c t γ = t = «=»=. 0.0 Q P to give v = 0.95c γ =.5 «length of train according Q» = 5/.5 «giving 00 m» d i Award [] for one gradient correct and another approximately correct. axes drawn with correct gradients of 5 for ct and 0.6 for x

8 8 M7/4/PHYSI/SP/ENG/TZ/XX/M 4 d ii lines parallel to the x axis and passing through B and F intersections on the ct axis at B and F shown light at the front of the train must have been turned on first d iii t = s Allow ECF for gamma from (c). d iv = 0000 according to P: ( ) according to Q: ( ) = 0000 e u = c = 0.9c

9 9 M7/4/PHYSI/SP/ENG/TZ/XX/M Option B Engineering physics 5 a i M vr a ii evidence of use of: M L = = MR + R I ω ( ) ω a iii evidence of use of conservation of angular momentum, v «rearranging to get ω =» 4R MvR 4 = MR ω a iv Mv initial KE = 6 Mv final KE = 4 energy loss Mv = 8

10 0 M7/4/PHYSI/SP/ENG/TZ/XX/M 5 b i Γ 0.0 α «=» = 4 MR «to giveα = rads» Working OR answer to at least SF must be shown b ii ωi θ = «from ωf = ωi + αθ» α v. θ «= =» =.8 OR.9 «rad» R α number of rotations «=» =.0 revolutions π

11 M7/4/PHYSI/SP/ENG/TZ/XX/M 6 a «a process in which there is» no thermal energy transferred between the system and the surroundings b A to B AND C to D c i PV T = nr T = 49 «K» The first mark is for rearranging. c ii PV a A PB = V P B = B 67 kpa The first mark is for rearranging. d i «B to C adiabatic so» 5 5 B B PV C C PV = AND PCV C = nrtc «combining to get result» It is essential to see these relations to award the mark. d ii T C 5 = PV B B V nr C (.0 0 ) T C = «(.90 0 )» = 4 «K» e the isothermal processes would have to be conducted very slowly / OWTTE

12 M7/4/PHYSI/SP/ENG/TZ/XX/M Option C Imaging 7 a i an image formed by extensions of rays, not rays themselves OR an image that cannot be projected on a screen a ii v = «v = cm» a iii u = 8 = 6.0 «cm» v = 4 «cm» «=» f = 8.0 «cm» f Award [ max] for answer of 4.8 cm. Minus sign required for MP. a iv line parallel to principal axis from intermediate image meeting eyepiece lens at P line from arrow of final image to P intersecting principal axis at F

13 M7/4/PHYSI/SP/ENG/TZ/XX/M 7 b i object is far away so intermediate image forms at focal plane of objective for final image at infinity object must also be at focal point of eyepiece «hence 87.5 cm» No mark for simple addition of focal lengths without explanation. b ii angular magnification = 85.0 = 4.50 angular diameter = rad c chromatic aberration is the dependence of refractive index on wavelength but mirrors rely on reflection OR mirrors do not involve refraction «so do not suffer chromatic aberration»

14 4 M7/4/PHYSI/SP/ENG/TZ/XX/M 8 a i longer distance without amplification signal cannot easily be interfered with less noise no cross talk higher data transfer rate max a ii infrared radiation suffers lower attenuation b.4 loss = 0log = db length = «=» «km» 0.0 c a thin core means that rays follow essentially the same path / OWTTE and so waveguide (modal) dispersion is minimal / OWTTE

15 5 M7/4/PHYSI/SP/ENG/TZ/XX/M Option D Astrophysics 9 a i stars fusing hydrogen «into helium» a ii 5.5 M = M (4 0 ) = 9.86M «M 40M» a iii 4 5 T 4 0 = T 4000 «K» a iv 5 AU 4 0 = 4 0 d 8 d = 0 «AU» Accept reverse working. Accept use of substituted 4 values into L= σ 4π RT. Award [] for a bald correct answer. Accept use of correct values L into b =. 4 π d b the gravitation «pressure» is balanced by radiation «pressure» that is created by the production of energy due to fusion in the core / OWTTE Award [ max] if pressure and force is inappropriately mixed in the answer. Award [ max] for unexplained "hydrostatic equilibrium is reached".

16 6 M7/4/PHYSI/SP/ENG/TZ/XX/M 9 c the Sun will evolve to become a red giant whereas Theta Orionis will become a red super giant the Sun will explode as a planetary nebula whereas Theta Orionis will explode as a supernova the Sun will end up as a white dwarf whereas Theta Orionis as a neutron star/black hole

17 7 M7/4/PHYSI/SP/ENG/TZ/XX/M 0 a black body radiation / K highly isotropic / uniform throughout OR filling the universe Do not accept: CMB provides evidence for the Big Bang model. a ii.9 0 «λ =».0 «mm».8 b the universe is expanding and so the wavelength of the CMB in the past was much smaller indicating a very high temperature at the beginning c i v 5 5 «z =» v = «= kms» c 5 v «d = v = = 706» 70 «Mpc» H 68 0 Award [ max] for POT error. c ii R R z =.6 R R = 0 0 R R =