GCE. Physics B (Advancing Physics) Mark Scheme for January Advanced GCE Unit G495: Field and Particle Pictures

Transcription

1 GCE Physics B (Advancing Physics) Mark Scheme for January 202 Advanced GCE Unit G495: Field and Particle Pictures Oxford Cambridge and RSA Examinations

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3 G495 Mark Scheme January 202 Annotations available in Scoris Annotation Meaning Benefit of Doubt given Contradiction Incorrect Response Error Carried Forward Follow through Not answered question Benefit of doubt not given Power of 0 error Omission Mark Rounding Error Error in number of Significant figures Correct Response Arithmetic error Wrong physics equation

4 G495 Mark Scheme January 202 Annotations in detailed mark scheme Annotation Meaning / alternative and acceptable answers for the same marking point () Separates marking points reject Answers which are not worthy of credit not Answers which are not worthy of credit IGNORE Statements which are irrelevant ALLOW Answers that can be accepted ( ) Words which are not essential to gain credit Underlined words must be present in answer to score a mark ecf Error carried forward AW Alternative Wording ORA Or reverse argument 2

5 G495 Mark Scheme January 202 (a) Wb m -2 (b) J C - Wb s - 2 (a) E = m 2 v 2 /2m shown Allow: any consistent method Allow: reverse argument (b) E k = h 2 /2m 2 shown Allow: any consistent method Allow: reverse argument (c) wavelength = 3.5 x 0-0 m 3 (a) C (b) A 4 E = 3000/4 x 0-4 = 7.5 x 0 6 (V m - ) positron/anti-electron () 6 Curving upwards between plates, straight (by eye) outside plates 7 (a) Same/ equal Allow: AW (b) Half, /2 Allow: Y is twice X Not: less -4.x 0^-9t 8 35 = 25 x e ln 35/25 = -4. x 0-9 t t = 3. x 0 8 s (= 9.7 years) 9 (a) C (b) r = 7 x 0-5 x / = 7 x 0-5 x 0.27 r =.9 x 0-5 m Allow: beta-plus Allow: ecf from incorrect numbers Allow: positron etc as independent mark Use Scoris marking tool if in doubt. Total 20 Allow:.84 half lives () so 9.7 years() Allow: Approx two half lives () so 0.6 years () Allow: Substitution () leading to 33.7kBq () after ten years Own value needed for second mark Or: r 0 =.2 x 0-5 m () Need own value 3

6 G495 Mark Scheme January (a) Complete loop within iron Loop will shorten when coil moves anticlockwise Allow: straighten, AW Ignore: reference to N and S attracting (b) (i) Iron has high (relative) permeability Not: permeance Allow: iron is a good conductor of flux Curved poles decrease air gap Flux is therefore increased Allow: curved faces provide for constant force/torque on rotor Allow: flux density/field strength increases (ii) Laminations limit/ reduce eddy currents Eddy currents oppose (existing) flux (c) Rate of change of flux (linkage) increases so induced emf increases in opposition to current (from supply) Not: just eddy currents reduce flux Not: flux reduction as a consequence of eddy current heating Allow: eddy currents set up opposing flux Allow: rate of cutting flux increases Allow: answers in terms of back emf that demonstrate clear understanding of cause. Total 0 4

7 G495 Mark Scheme January 202 (a) Radial field Allow: any four correct lines Arrows pointing outwards (b) V = 9 x 0 9 x 2.5 x 0-9 /4 x 0-3 = 5625 V Need own value (c) (i) Taking pairs of points, multiply x and y value values will be constant (ii) E = V/r = 2800/ = 350 x 0 3 NC - Allow: replot data as V against /r () and showing straight line through origin () Allow: V halves as r doubles () for more than one pair of points () Not: V decreases as r increases Not: replot y against /x Method must clearly relate to radial field Or: gradient of graph at m () calculation to value in range 300 to 400 knc - () Not: E = V/d (0 marks) (d) Charges will migrate towards outer edges of sphere increasing distance between the centres of charge Total 0 5

8 G495 Mark Scheme January (a) anti-(electron) neutrino (b) i change of mass =.42 x 0-3 u = 2.36 x 0-30 kg Energy release = 2.2 x 0-3 J =.33 Mev Allow: ecf from any stage to next (ii) (Kinetic) energy taken away by (anti) neutrino Allow: ecf for particle name only Not: energy to create antineutrino Not: heat/light (any mention = 0 marks) (c) (i) activity = x 2.2 x 0 20 /4.2 x 0 6 = 3.6 kbq (ii) dose = 3600 x 7 x 0-4 x 3.2 x 0 7 /60 =.3 x 0-4 Gy Assumptions clearly stated e.g. All beta particles/energy absorbed by body; no other beta emitters present; level of potassium in body remains constant (iii) risk =.3 x 0-4 x 0.03 x 50 x 40 x 0 6 = 800 Or: =.65 x 0-7 () Must have own answer Allow: 3.7kBq Allow:.3 to.5 x 0-4 Gy Not: constant rate of decay Allow: 7800 to 9000 Allow: ecf from (c) (ii) POT error max Total 3 6

9 G495 Mark Scheme January (a) (i) To stop collisions (taking energy away from protons) Allow: AW (ii) Force on protons from B -field is at right angles to their direction of motion (b) 400 x.6 x 0-9 x 50 = 9.6 x 0-5 J (c) r = (2 x.7 x 0-27 x 9.6 x 0-5 ) /2 /(0.8 x.6 x 0-9 ) = m Method mark Allow: 0.04 Not: or 0.05 (rounding errors) (d) Any three from: Gamma factor = (rest energy + kinetic energy)/rest energy Kinetic energy of both particles the same but the proton rest energy is much greater so gamma factor is larger for electron Calculation of gamma factors: electron =.2 Calculation of gamma factors: proton =.00 Larger gamma factor for electron means more relativistic behaviour 3 Allow: Electron faster for same KE () Gamma factor is larger explained using formula () First marking point can be implicit in correct calculations Last marking point dependent on previous argument Total 9 7

10 G495 Mark Scheme January (a) Use of distance = (50/60) x 400 = 000 m. Method mark. Allow: reverse argument (b) speed is not constant () One from: because it accelerates at the start () because buoyancy not constant () because drag varies () because density of atmosphere varies () because of vertical component of wind () Allow: other reasonable answers Not: variation of g with altitude Not: just wind OR: difficult to spot the exact moment of disappearance () AW because clouds are diffuse () AW (c) Use of pv/t = constant Giving V = m 3 Or: correct use of pv = nrt or pv = NkT Allow: Not: 0.032, Correct bald answer scores both marks Total 5 8

11 G495 Mark Scheme January (a) (i) & (ii) Any two pairs from Low density () reduces mass/ weight () 4 Reason must relate to the property Not: light for property but BOD for reason OR Low stiffness/ low Young Modulus () Makes balloon flexible/easy to inflate () Not: flexible for property but BOD for reason. Not: elastic OR Strong/ sufficient breaking stress () Balloon must not burst too soon () Not: to lift heavy loads OR Tough/ Not brittle () Fabric must not tear/split too soon/when cold () (b) Use of circumference = x D to establish increase in length = 6 or 8.8m strain = x/l = 8.8 / 6.28 = 3 or 300% Allow: (8-2)/2 = 3 for second mark only (c) V = 4/3 r 3 = 4.9 m 3 m = V = 5.03 kg B = 5.03 x 9.8 = 49.3 N Allow: ecf from any stage to next Allow: 49, N (d) Resultant force = 49.3 (.25 x 9.8) = 37.0 N a = F / m = 29.6 m s -2 Not: 39.4ms ms -2 (0 marks) Allow: 30.2 from 50N, 29.4 from 49N 9

12 G495 Mark Scheme January 202 (e) Examples: 3 Balloon expands as it rises () so buoyancy increases () so larger acceleration () Balloon expands at it rises () so drag force increases () so smaller acceleration () Allow: reasonable alternatives QWC: Poorly marshalled arguments cannot lead to more than 2 marks. Air density decreases with altitude () so buoyancy decreases () so smaller acceleration () Air density decreases with altitude () so drag decreases () so larger acceleration () Balloon is accelerating () so drag force increases () so acceleration decreases () Colder with altitude () so volume/buoyancy reduces () so acceleration reduces () Atmospheric pressure decreases with altitude () so volume/buoyancy increases () so acceleration increases () Total 4 0

13 G495 Mark Scheme January (a) 00 m / 4352 pixels = 0.25 m / pixel Allow: Not: 0.252, 0.3 (b) (i) 8 bits required (per colour pixel) bits = 4352 x 3264 x 3 x 8 = 3.4 x 0 8 bits (ii) 3.4 x 0 8 bits = G bytes () => 8 Gbytes / = 87.8, so 87 pictures () 2 Allow: ecf from (b)(i) Not: 88 OR 8 Gbytes = 6.4 x 0 0 bits () 6.4 x 0 0 bits/3.4 x 0 8 bits = 87.8 so 87 pictures () Total 5 7 (a) v = (2E/m) ½ Calculation in (M)eV = 0 marks = 4.3 x 0 7 m s - POT error - e.g. 4.3 x 0 4 for missing M (b) v = d/t = 3.5 x 0 3 / 0.3 x0-3 = 2.69 x 0 8 ms - Then, v/c = 2.69/3.0 = i.e 90% (c) / ( ) -½ (=2.26 or 2.29) t = 0.3 x 0-3 /( ) -½ => t = 5.7 x 0-5 s Total 7 Allow: reverse argument Needs own value or clear method Or: Recognition that t = t or t = t/ ()

14 G495 Mark Scheme January (a) E = kt = h c / => = h c / k T = 5.2 x 0-3 m Or: E = kt = 3.8 x 0-23 () Allow: use of 3kT/2 for full credit ( E = 5.7 x 0-23, = 3.5 x 0-3 ) (b) Range = 600 K Then, 600 K / 2.7 K = 2.2 x 0-4 = 0.022% Need own value or clear working (c) Reference to early universe mass concentrations/density variation/coalescence of matter/gravity variation formation of galaxies AW Allow: stars Total 7 QWC: three marks only awarded if answer is carefully ordered and clear 2

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