Oxford Cambridge and RSA GCE Physics B (Advancing Physics) Unit G494: Rise and Fall of the Clockwork Universe Advanced GCE Mark Scheme for June 05 Oxford Cambridge and RSA Examinations
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G494 Mark Scheme June 05 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 or equation
G494 Mark Scheme June 05 Abbreviations, annotations and conventions used in the detailed Mark Scheme (to include abbreviations and subject-specific conventions). Annotation Meaning / alternative and acceptable answers for the same marking point () Separates marking points reject Answers which are not worthy of credit not IGNE ALLOW Answers which are not worthy of credit Statements which are irrelevant Answers that can be accepted ( ) Words which are not essential to gain credit ecf AW A Underlined words must be present in answer to score a mark Error carried forward Alternative wording Or reverse argument The following questions should be annotated with ticks to show where marks have been awarded in the body of the text: Unless stated otherwise in the mark scheme, accept calculations which round to the mark scheme answer for full marks. 4
G494 Mark Scheme June 05 a N kg - b N m kg - EITHER ignore references to expansion of the universe red shift of light/radiation from (distant) galaxies; accept increase/stretching of wavelength as red shift because galaxies are moving away from each other / have recessional velocity owtte ; (uniform) microwave background; which is red-shifted light from early universe; accept galaxies moving away from Earth accept cosmological microwaves accept radiation for light γ = 4/8 =.; v = c γ look for evidence of correct transposition of data sheet formula 8 8 v =.0 0 =.7 0 ms 5.44 no ecf for incorrect γ 4 a initial KE = 50 00 =.00 0 J ; accept x0 6 J 5 final KE = (50 + 50) 5 =.5 0 J.x0 5 J; inital momentum: 50 x 00 =.0 0 4 kg m s -, final momentum = (50 + 50) x 5 =.0 0 4 kg m s - look for full working in calculations, not just final value accept 400 as final mass if 50 + 50 shown elsewhere look for words like initial/before and final/after as labels to calculations of energy or momentum accept alphabetic suffixes e.g. i, b, f or a to p and E as labels work done deforming the spacecraft; accept transfer to heat or thermal energy, ignore sound 5 b C 6 a -0.(0); -0.(0); (0.05-0.05 0.45 =) -0.0075-0.008 6 b any of the following: formula has effectively infinite number of steps not enough steps in the iterative calculation time interval too long in iterative calculation each iteration assumes constant speed allow ecf if v = incorrect Δv no ecf for incorrect v accept step-widths of zero time ignore constant acceleration other than zero acceleration 5
G494 Mark Scheme June 05 7 evidence of suitable test e.g. is ρt constant; test applied to all data to s.f. or s.f.; 7 x.9 = 5 (50) 8 x.5 = 54 (50) 9 x.0 = 5 (50) 0 x.6 = 5 (50) accept calculations to more than s.f with a conclusion which mentions that numbers are either not the same to s.f. or the same to s.f. ignore any conclusion about the truth of the relationship 8 EITHER ln measure time for half the sample to decay (t / ),use λ = t ; / take a known number of atoms, measure activity and use equation given; measure gradient (-λ) of a ln(activity)-time graph; accept measure half-life from activity-time graph 9 B Section A Total 0 6
G494 Mark Scheme June 05 0 a c mv GMm mv mv GMm = ; accept mr ω as not = r r r r r π r evidence of v = ; π accept ω =, v = rω, ω = πf T T algebraic manipulation to final formula; look for clear steps ignore loss of minus sign in final manipulation 4 look for correct formula or evaluation V = π r =.098 0 m ; 4 M = ρ V =.7 0.098 0 =.96 0 kg accept.0x0 4 with full working for [] accept V =.x0 m gives.97x0 4 kg for [] any two points from: time for pulse to reach moon = time for pulse to not distance for time, accept (laser) light for pulse return; radius of Moon / Earth is comparably negligible; speed of pulse is constant (throughout the journey) or Earth's atmosphere does not affect speed of not just travels at the speed of light pulse or pulse travels at speed of light in a vacuum (x0 8 m s - ) or speed of light is constant ; b i ii iii 8.0 0.5 8 r = =.75 0 m ; 8 4 π (.75 0 ) 0 G = =. 0 N m kg 4 6.0 0 ; (.4 0 ) density / mass of Earth incorrect; need to use density of whole Earth / core and mantle are made of different material / density increases with increasing depth; Total no ecf for incorrect value of r r =.8x0 8 m gives G =.5x0-0 or.x0-0 accept mass / density is too great accept orbit may not be circular for [] ignore references to radius of Earth and Moon 7
G494 Mark Scheme June 05 a molecules bounce off the ground; any two of the following. each bounce transfers momentum to ground accept collide with the ground accept impulse as momentum transfer / change. force on ground is rate of transfer of momentum Δ p F. pressure is force per unit area ignore algebraic formulae e.g. F =, P =, Δ p = mv mu Δ t A QWC: first marking point i NkT = Nmc ; not kt = mc b T = 9 K; kt c = = 5 ms 50 m s - m ecf T:= 0 K gives 4 ms - for [] 5 allow c =.6 0 for [] any one of the following assumptions. elastic collisions ii. molecules impact surface at right angles to it. all molecules moving at rms speed evidence of use of F = PA (=.0x0 5 x 0.56 = 5.6 0 4 N); EITHER Δ p nm c F = = and n =. 0 7 s - ; () Δ t Δ p nm c F = = and n =.x0 7 s - ; () Δ t Total 0 c = 500 m s - gives.4x0 7 s - 8
G494 Mark Scheme June 05 a i V 6.0 ( R = = ) =.() 0 Ω I.7 0 EITHER use of half-life = 0.69RC = ± s / 5± s; ii use of RC = /e drop time = 49± s / 69± s; t / RC data from graph, use of I = I 0 e ; evidence of method [] C =.±0. 0 - F correct answer [] b i 6 5 Δ Q = C ΔV = 470 0 0. = 5.64 0 C ; 5 Δ Q 5.64 0 7 I = = = 9.4 0 A Δ t 60 = 0.94 μa ; no ecf on incorrect ΔQ ii energy required; for an electron to break free (from an atom) / enter the not just to get from one plate to the other conduction band / become a free electron / move freely within the insulator; ε look for method which will eliminate A [] use of ln I = ln A- to eliminate A; kt iii ε =.9 0-0 J; corrrect answer for [] Total 9 9
G494 Mark Scheme June 05 a large amplitude (vertical) oscillations; make it dangerous/unpleasant for occupants of lift ; b i F use of k = or F = kx; x use of x E A L ii 4.0 0.5 0 5 k = =.67 0 Nm 00 m 500 + 640 T = ( π = π ) = 0.7s 5 k.67 0 f = = =.4 Hz T 0.7 c i idea that damping requires friction / energy transfer from lift AND slowing down the lift / reducing efficiency of lift EITHER reducing L (to increase k); ii raising f 0 (above Hz); increasing mass of load / cage; lowering f 0 (below 0. Hz); decreasing mass of load / cage raising f 0 (above Hz) increasing csa of cables (to increase k) raising f 0 (above Hz); use a cable material which is stiffer / increased E; raising f 0 (above Hz) Total 0 ignore references to sideways oscillations / swinging accept break cables F accept k = Δ L no ecf on incorrect k allow ecf if mass is just 500 kg (.7 Hz) or 640 kg (.6 Hz) for [] any realistic modification [] which is explained [] not increasing L not reducing csa of cables not more elastic material QWC against second marking point (organise information clearly) 0
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