ACSI 2024 Promo HL Phys MS

Y5 IBDP EOY HL Physics Examination 2024 Marking Scheme PHYSICS P1a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 C B C D A D D B A B D C C D B C B A C B 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 B A B B B D B D D C C B D B C D A D C B PHYSICS P1b Answer Marks 1(a)(i) uncertainty in period reduces; (you can show that % uncertainty in T reduces from 8.7% to 0.43%, a reduction of about 20 times) by 20 times; improves the precision of period; 2 1(a)(ii) Human reaction time in using stopwatch; Likely to be at least 1 10 th a second (0.1 s); value student recorded is too precise; 2 1(a)(iii) T = 0.5750 ± 0.0025 s; T2 = 0.3306 s2 % uncertainty in T = 0.435 % Hence % uncertainty in T2 = 0.870 %; T2 = 0.870% of 0.57502 = ± 0.0029 = ± 0.0003; 3 1(a)(iv) % uncertainty in L= % uncertainty in F + 2(% uncertainty in T) + % uncertainty in m = 5.88 + 0.8696 + 0.5; = 7.3 %; This question has an ecf full marks from 1(a)(iii). 2 1(b) i Plot T2 (y-axis) vs m (x-axis) or T (y-axis) vs √𝑚(x-axis); 1 1(b) (ii) <m> = 202 g (± 1 g) m = ± ½ (210 – 195) = ±7.5 g = ± 8 g 2 2(a)(i) The effective acceleration along the slope is equal to 𝑔 = 𝑎 sin 𝜃 or 𝑎 = 𝑔 𝑠𝑖𝑛 𝜃 1 2(a)(ii) t = 0.4 s, x = 0.2  1002 cm hence g = 2x / t2 = 250 cm s-2 (Choose at least greater than 0.35 s (half the graph) and easy and accurate to read the value of x.) Hence g = a ÷ sin 15 = 970 cm s-2 ; This question has an ecf full marks from 2(a)(i). 3 2(a)(iii) human reaction in the time taken; take several repeats of the same measurement; determine average; Alterative answer: direction of wind is random and so time taken is either lower or higher than true value; minimize this random error by taking several repeats of the same measurement and determine average; 2 2(b) (smaller acceleration) hence larger time; (% / fractional) uncertainty of time decreases; 2

PHYSICS P2 Answer Marks 1(a) Using s = ut + ½ at2 y =½ (9.8) t2 100 = ½ (9.8) t2 t = 4.5 s 1 1 1 1(b) 1 mark for each correct graph 1 1 1(c) Total horizontal displacement travelled by parcel = 72 × 4.5 = 324 m Speed of truck = 324 125 444.5 − = m s−1 (if student uses 4.52 s from part (a), then the answer could be 44.5 or 45 ms-1) 1 1 2(a) Correct labelling and directions of forces. Correct relative magnitude of forces (vertical components). Minus 1 if relative magnitude of vertical components is obviously not correct. 1 1 2 (b) o4.00 2.50 sin 28.0 5.2 m= + =r 1 2 (c) 1 x/ m t/ s 125 4.5 T B Tension Weight

2 2 T cos θ= mg mvT sin θ= r vtan θ = rg o(tan 28.0 )2v = r g o5.17(9.81)( tan 28.0 ) 5.2 m==v s−1 1 1 2 (d) 50 9.8 cos 28 560 N = = T cos θ= mg mgT= cos θ 1 1 3 (a) F = kx = 75 × 0.085 = 6.38 N 1 1 3 (b) mgh = ½ kx2 or ½ Fx 0.025 × 9.81 × h = ½ × 75 × 0.0852 height h = 1.1 m 1 1 3 (c) The maximum height is reduced / lower. Energy is lost to work done against air resistance / resistive forces 1 1 4a(i) Heat lost by iron ball = Heat gained by copper calorimeter and water 24 0.451 (T-22) = (40  0.385  7) + (300  4.2  7); T = 847 oC = 850 oC (2 sfs) 1 1 4a(ii) Loss in the mass of water (due to rapid boiling occurring as the temperature of iron ball is much greater than the boiling point of water.) (The temperature T will be lower.) Do not accept thermal energy losses by the iron water during the transfer or thermal energy losses by the water and calorimeter to the surroundings 1 4b by energy balance model, I1 = I + I2 = 390 W m-2; I1 = T4 , hence T = √ 390 5.67×10−8 4 = 290 K (2 sfs); (also accept 288 K) 1 1 4c Use 𝑝𝑉 = 𝑛𝑅𝑇; from graph 𝑝𝑉 = 12; Hence 𝑇 = 12 4.5×10−3×8.31 T = 320 K (2 sfs); 1 1 5a(i) Vlamp = IR = 0.30  5.0 = 1.5 V; hence V3.0 = 4.5 V; I3.0 = 4.5 ÷ 3.0 = 1.5 A R = 1.5 1.5−0.30 = 1.5 1.20 = 1.25 Ω ; 1 1 5a(ii) Total power is given by 𝑃 = 𝑉𝐼 = (6.0)(1.5) = 120 𝑡 ; giving 𝑡 = 13.33 s; Hence 𝐸𝑙𝑎𝑚𝑝 = 𝐼2𝑅𝑡 = 0.302 × 5 × 13.33 = 6.0 J; 1 1 1

5a(iii) 𝑉 = 𝐼𝑅 6.0 = 1.176 (r + 3 + 1.25); r = 0.85  (2 sfs); 1 1 6ai 1) acceleration is proportional to Displacement / vice versa Because graph is straight line passing through origin 2) Displacement and acceleration are opposite in direction Acc. Towards equilibrium Also accept constant and negative gradient (makes reference to a = -2 x) 1 1 aii (for show question, student needs to show raw value of f as 349 Hz) 1 1 bi = 𝐼 × 𝐴 𝑃 = (1.0 × 10−2)(5.0 × 10−5) = 5.0 × 10−7𝑊 𝐸 = 𝑃𝑡 𝐸 = (5.0 × 10−7)(5)(60) = 1.5 × 10−4 𝐽 1 1 bii No energy is lost to the surroundings/ OR all the energy is transmitted from the monitor to the eardrum. Accept sound energy is incident at right angle to the plane of the eardrum 1 6c i T = 0.44 s 1 cii 𝜔 = 2𝜋 𝑇 = 2𝜋 0.44 = 14.3 𝑟𝑎𝑑 1 2 𝑚𝜔2𝑥02 = 10 𝑥0 = √ 2(10) (0.300)(14.3)2 = 0.57 𝑚 1 1 1 2

6ciii 1 mark for shape. 1 mark for proper axis labels and labelling of pertinent displacement and time values. 7ai 3 7aii Loud and soft sound is heard- accepted 2 0.22 Displacement/ m Time/s 0.44 0.57

7aiii Allow 𝐼 ∝ 𝐴2 for 2nd mark 2 7bi d = 1 8.00 𝑥 105 = 1.25 x 10-6 m d sin θ = nλ θ = sin-1 [(2 𝑥 589 𝑛𝑚)/(1.25 𝑥10−6) = 70.459 θ = sin-1 [(2 𝑥 590 𝑛𝑚)/(1.25 𝑥10−6) = 70.734  = 0.275 OR Ans in rad is accepted with working  = 1.23455- 1.22973 = 4.81 x10-3 radians 1 1 1 1 4 7bii The lines are closer together / not clearly separate in the first order spectrum hence not resolved. 1 8a(i) I = m r2; where m is the mass of the flywheel and r is the radius of the flywheel 2 8a(ii) f = I + t 300 = 0 + (150);  = 2.0 rad s-2 ; Hence  = I   = 90  2.0 = 180 N m; 1 1 1 8a(iii) Erot = ½ I 2 = ½ (90)(300)2; = 4.1 106 J ; 1 1 8a(iv) 𝑊 = 𝐹 × 𝑑 4.05  106 = 400  d d = 1.0 104 m ; 1 8b(i) The net torque acting on the system must be zero. 1 8b(ii) 90 rev/min → f = 1.5 Hz i = 2f = 3 Similarly f = 2(1.33) = 2.66 ; Conservation of angular momentum (Idisc)(3) = (Idisc + 0.02  0.052)(2.66) ; Solving, Idisc = 3.9  10-4 kg m2 ; 1 1 1 9a(i) 𝑝𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑝2 = 4.0×105×3.0 5.0 𝑝2= 2.4 × 105 Pa; 1 9a(ii) Sketch shows the curve starts at (3.0 m3, 4.0 105 Pa) and ends at (5.0 m3, 2.4 105 Pa) and correct shape of the curve 2 1

9a(iii) W = area under the curve  area of trapezium = 1 2 (6.4 × 105)(2.0); W = 6.4  105 J; (also accepts exact value using 𝑊 = 𝑛𝑅𝑇 𝑙𝑛 𝑉2 𝑉1 = (4.0105) (3.0) 𝑙𝑛 5 3 = 6.1  105 J) 1 1 9a(iv) Less work is done because the adiabatic change will have the curve that is below the curve in part (a)(ii) and so the area under the curve will be smaller OR the curve representing the adiabatic change will be steeper than the isothermal change. 1 9b(i) W = area of the rectangle = (2  105) (8.0) ; W = 1.6  106 J; 1 1 9b(ii) Efficiency = 𝑊 𝑄𝑖𝑛𝑝𝑢𝑡 = 1.6×106 (1.8+1.6)×106 ; = 0.47 = 47% (2 sfs); 1 1 9c The second law of thermodynamics states that the total entropy of a system either increases or remains constant in any reversible process; it never decreases. 1 10 a . 3 10 b 𝑓′ = 𝑓 𝑣 𝑣 − 𝑢𝑠 = 293 = 275 330 330 − 𝑢𝑠 Speed of the source = 20.28 = 20.3 ms-1 2