A Level Physics Paper 1 Practice Paper Mark Scheme Summer 2022

Document Content and Description Below

A Level Physics Paper 1 Practice Paper Mark Scheme Summer 2022 Page 1 Section A 1 (a) (i) any two eg proton, neutron 2 (ii) 1 (b) (i) contains a strange quark 1 or longer half life than expected... or decays by weak interaction (ii) the second one because lepton number is not conserved 2 lepton number LHS = 0 RHS -1 + -1 = -2 (c) (i) weak (interaction) 1 (ii) X must be a baryon 2 baryon number on right hand side is +1 [Total 9 marks] 2 (a) (i) (moment = 520 x 0.26) = 140 (135.2) Nm 2 (ii) 180 x 0.41 and 0.63 X seen 3 135.2 = 180 x 0.41 + 0.63 X ecf from (a)(i) (X = (135.2 – 73.8) / 0.63) = 97 (N) (97.46) allow 105 from use of 140Nm ecf from (a)(i) (iii) (520 – (180 + 97.46)) 1 = 240 (242.5 N) ecf (or from correct moments calculation) [Total 6 marks] A-LEVEL PHYSICS MARK SCHEME Practice Paper 1 Maximum marks: 85 View detailed guidance on the conclusions you can draw from your students' performance in these papers on the MERiT welcome page. Understand how your students compare with others and target revision effectively by entering marks into MERiT. This document is licensed to Haggerston School - MB260045 Page 2 3 (a) (momentum of air) increases ✔ 1 words implying increase (b) (rate of change of momentum so) force acting on the air (Newton 2) ✔ 3 it/air exerts force (on engine) of the same/equal magnitude/size ✔ but opposite in direction (Newton 3) ✔ allow backwards and forwards to indicate opposite (c) (use of F = Δmv/t) 1 F = 210 × 570 = 120 000 (N) (119 700) ✔ (d) Momentum/velocity is a vector ✔ 2 Momentum/velocity has a direction ✔ There is a change of direction ✔ A direction change is a momentum change ✔ Any of above points - max 2 marks. (e) use of v2 = u2 +2as 2 forgetting to square u or double a score 1 mark 0 = 682 − 2 × 2.7 × s ✔ s = 682/(2 × 2.7) = 860 (m) (856) ✔ accept range 850 − 860 accept alternative method first mark for time calculation AND correct substitution second mark using s = ut +1/2at2 OR average speed (f) rate of intake of air decreases (as plane slows) OR volume/mass/amount 2 of air (passing through engine) per second decreases ✔ allow argument in terms of (air) resistance for 1 mark only i.e air resistance decreases as speed of aircraft decreases (as) smaller rate of change of momentum OR momentum change ✔ NOT FRICTION [Total 11 marks] Q4 (a) (current in) steel wire (is less than the current in an) aluminium wire as it 3 has a higher resistivity / resistance OR aluminium is better conductor✓ the six aluminium wires are in parallel OR total cross-sectional area of aluminium is 6 times greater than steel wire✓ each aluminium wire carries three times as much current as the (single) steel wire✓ (b) If the steel wire is ignored then can score first and third marks 3 For correct answer accept range 32 kW to 33 kW If wires treated as series resistors then zero marks This document is licensed to Haggerston School - MB260045 Page 3 Alternative methods resistance of 1 km of 6 Al cables in parallel = = 0.183 Ω✓ total resistance of the cable = 0.174 Ω✓ power loss per km = 32.3 kW (or 30.7 kW if they ignore the steel)✓ OR power loss in 1 km of steel = 1.70kW✓ power loss in 1 km each of Al cable = 5.11 kW✓ total power loss per km = 32.4 kW (or 30.7 kW if they ignore the steel)✓ OR calculate current in steel wire and aluminium wire (22.7 and 68.2) ✓ calculate power loss in steel wire & 1 aluminium wire (1700 and 5115) ✓ calculate total power loss (1700 + 6 × 5115 = 32,4 kW) ✓ [Total 6 marks] 5 (a) correct substitution into P = V 2/R ✔ 5 (condone power of 10 error) R = 2.62 (Ω) = 144/55 = 122/55✔ correct substitution into ρ= RA/L ✔ (condone error on R and/or power of 10 errors) resistivity = 9.9(5) × 10–7 (range 9.9 to 9.95 × 10–7) ✔ unit = Ω m✔ (b) lost volts = 0.1 (V) or 0.1 seen as voltage✔ 2 r = 0.011 to 1.09 × 10–2 (Ω) ✔ (c) brightness decreases✔ 3 increased current (in circuit/battery) ✔ increased lost volts leading to decreased pd across bulb or decreased terminal pd✔ [Total 10 marks] 6 (a) f = 1/T =1/0.05=20 Hz ✔ 2 A = ✔ (=3.5 × 10−4 m) (b) Cosine shape drawn, maximum at t= 0, amplitude 3.5 × 10−4 m ✔ 1 (c) (any of the following when the velocity is zero) 1 0.00s, 0.025s,0.050s or 0.075s ✔ (d) when the vibrating surface accelerates down with an acceleration less than 3 the acceleration of free fall the sand stays in contact. ✔ above a particular frequency, the acceleration is greater than g ✔ there is no contact force on the sand OR sand no longer in contact when downwards acceleration of plate is greater than acceleration of sand due to gravity ✔ (e) (when the surface acceleration is the same as free fall) 2 g = Aω2 = A (2 πf)2 ✔ f = √(g /A4 π2 ) = (9.81/(3.5 × 10−4 × 4 π2))1/2 = 26.6(7) Hz ✔ [Total 9 marks] This document is licensed to Haggerston School - MB260045 Page 4 7 (a) (i) absorbs enough energy (from the incident) electron (by collision) 2 OR incident electron loses energy (to orbital electron) ✔ exact energy / 10.19(eV) needed to make the transition / move up to level 2 ✔ For second mark must imply exact energy Energy required to move from level 1 to level 2 (ii) (use of E2 –E1) = hf 3 −3.41 − − 13.6 = 10.19 ✔ energy of photon = 10.19 × 1.6 × 10−19 = 1.63 × 10−18 (J) ✔ 6.63 × 10−34 × f = 1.63 × 10−18 f = 2.46 × 1015(Hz) ✔ (accept 2.5 but not 2.4) ECF from energy difference but not from energy conversion (iii) Ek = 1.7 × 10−18 − 1.63 × 10−18 2 = 7.0 × 10−20 J ✔ (b) Electrons return to lower levels by different routes / cascade / not straight 2 to ground state ✔ Three transitions stated n=3 to n=1 ; n=3 to n=2 ; n=2 to n=1 ✔ [Total 9 marks] This document is licensed to Haggerston School - MB260045 Page 5 Section B Question Key 8 B 9 A 10 C 11 D 12 D 13 A 14 C 15 A 16 C 17 B 18 A 19 B 20 A 21 C 22 D 23 B 24 D 25 D 26 A 27 B 28 C 29 C 30 C 31 D 32 B This document is licensed to Haggerston School - MB260045 [Show More]

Last updated: 3 years ago

Preview 1 out of 5 pages