GCE Physics B (Advancing Physics) Advanced GCE G494 Rise and Fall of the Clockwork Universe Mark Scheme for June 200 Oxford Cambridge and RSA Examinations
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G494 Mark Scheme June 200 Question expected answer mark Additional guidance a N kg - b J kg - K - 2a E = 0.5CV 2 =.2(5) 0-3 J ignore anything more than two sig. figs e.g.2 0-3 for [] 2b 3a 3b 3c (current is) flow of charge (off capacitor, through circuit); EITHER p.d. across capacitor/resistor decreases (as capacitor loses charge) rate of charge release proportional to charge (on capacitor) Nm 3p D = = ( ) V 2 c 2 3p c = = 500 m s - D gas particles change direction; because they collide (with other particles); ignore references to discharging not just charge on capacitor decreases accept electrons as charge accept I = Q/t or wtte for [] look for D = Nm/V N = earns [0] ignore references to random ignore description of random walk 4a 4b A.6 0 N = = = 2. 0 0 λ 7.6 0 A = A 0 e -λt = 4.7 0 4 Bq 5 4 N = 2. 0 4 gives 4.75 0 4 Bq for [] ignore sign of answer
G494 Mark Scheme June 200 5 EITHER C and D D and C 2 each correct response for [] remember Don't Care 6a 6b N pv = kt = 3. 0 NkT =.2 0 4 J 24 7 correct pattern for [] look for at least two sig. figs in their answer, rounding to 3. 0 24 accept correct reverse calculation 3NkT/2 gives.8 0 4 J for [] 5NkT/2 gives 2.9 0 4 J or 3.0 0 4 J for [] 8 correct pattern for [2] one mistake for [] a mistake is an extra tick, a missing tick or a tick in the wrong place 9 A C D 2 only one mistake for [] remember All Can Do 2
G494 Mark Scheme June 200 0a particles bounce off ground; momentum of particles/ground changes; EITHER force on ground is its rate of change of momentum momentum change of particle requires a force, so equal and opposite force on the ground accept collide/hit QWC mark - must use correct terms for third marking point 0bi p = (N/V)C where C = kt = constant 0bii probability of a particle at h is e ε kt where ε = mgh is gravitational/potential energy of particle; ; accept fraction/proportion for probablility accept GPE 0biii 7.9 0 4 Pa accept more than 2 sig. fig. 3
G494 Mark Scheme June 200 0c EITHER more energy in system so more particles get lucky and able to reach that height T = some value above 290 K (b)(iii) correctly recalculated increased KE/speed/velocity of particles increased collision rate increased momentum of particles increased momentum change per collision kt increases accept pv = nkt so p increases with increasing T for [] 2 accept increase of c so p = ρc / 3 increases for [] accept (-)E/kT or (-)mgh/kt decreases (as T increases) 2 so E kt mgh e e / BF / kt increases 4
G494 Mark Scheme June 200 ai all lines at 45 from (0,0) changes direction at (?,4) and ends at (0,8) accept lines drawn freehand aii time out = time back because speed of light is constant accept one-way time is half total time ignore references to distance accept pulse travels at speed of light aiii s = 3.00 0 8 4.00 =.20 0 9 m accept.2 0 9 m bi bii c pulse-echo time is now 7.34 s less than before, (so reduced distance) s at 950 s is.0 0 9 m (.20-.0) 0 9 /946 =. 0 5 m s - (measure) change of wavelength Δλ apply z = Δλ/λ = v/c accept pulse-echo time is reduced by 0.66 s for [2] accept calculation of new distance of. 0 9 m for [2] allow ecf from incorrect bi and aiii accept.(0) 0 5 m s - allow ecf incorrect new distance to asteroid for [] accept increase or decrease of wavelength accept measure the red/blue shift for [] 5
G494 Mark Scheme June 200 2ai Sound energy produced (at expense of kinetic energy). 2aii Δp = 2.0 (5 + 3.3) = 6.6 Ns for hammer p = 6.6 Ns for mass v = 6.6 / 0 =.7 m s - look for attempt at momentum conservation [] correct substitution for [] evaluation for [] so 0.34 m s - for [2] 2b correct shape and period correct phase over the whole time span, any constant amplitude 2ci a = -50 0.2 = - 0.5 m s -2 v = 0.85-0.5 0.05 = 0.325 m s - average speed = 0.5875 m s - x = 0.2 + 0.59 0.05 = 0.24 2cii (do two or more successive calculations) for shorter time intervals allow ecf from one step to the next accept correct use of s = ut + at 2 /2 for full marks ignore use of x = Acos(2πft) 6
G494 Mark Scheme June 200 3a [] for each correct arrow 2 mark the direction of the arrow if it doesn't pass through the comet 3bi E k = /2mv 2 so E k /m = v 2 /2 E k /m = (54.6 0 3 ) 2 /2 =.49 0 9 J kg - 3bii GMm E g GM E g = so = ( =V g ) r m r = -6.67 0-2.00 0 30 / 8.82 0 0 = -.5() 0 9 J kg - E t = E g + E k = (-2 0 7 J kg - ) ignore calculation of total energy, but accept -.2 0 7 J kg - 3biii E g = - 2.5 0 7 J kg - E k =-2.0 0 7 -(-2.52 0 7 ) = 5.2 0 6 J kg - 6 so v = 2 5.2 0 = 3.2 0 3 m s - calculate new value for E g using -GM/r for [] ecf: calculate new E k by E t - E g for [] ecf: calculate v from E k for [] accept 2.3 0 3 m s - for [3] 7
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