DHS 2022 Prelim P2 Guide

1 2022 DHS H2 Physics Prelim Paper 2 Suggested Solutions 1 (a) vx = (6.02 – 4.82)1/2 M1 = 3.6 m s–1 A0 or 6.0 sin  = 4.8, so  = 53.1° and vx = 6.0 cos 53.1° M1 = 3.6 m s–1 A0 (b) (i) straight line from (0, 4.8) to (0.49, 0) A1 straight line continues with same slope to (0.98, –4.8), labelled Y A1 (ii) a horizontal line A1 from (0, 3.6) to (0.98, 3.6), labelled X A1 (c) s = ut + ½at2 = (4.8  0.49) + (½  –9.81  0.492) C1 = 1.2 m A1 or s = ½(u + v)t = ½  (4.8 + 0)  0.49 C1 = 1.2 m A1 or v2 = u2 + 2as s = 4.82 / (2  9.81) C1 = 1.2 m A1 Y X time / s

2 (d) kinetic energy at maximum height change in gravitational potential energy = 1 2⁄ (mvx2) mgh = 1 2⁄ (3.62) 9.81(1.2) C1 = 0.56 A1 or kinetic energy at maximum height change in gravitational potential energy = 1 2⁄ (mvx2) 1 2⁄ (mvY 2) = 1 2⁄ (3.62) 1 2⁄ (4.82) C1 = 0.56 A1 (e) With air resistance, the resultant force is larger than weight, resulting in larger deceleration, hence the actual time is shorter. B1 2 (a) energy (stored) / work done represented by area under graph B1 energy = ½ (180) (4.0 x 10-2) C1 = 3.6 J A1 (b) (i) either momentum before release is zero M1 so sum of momenta of trolleys after release is zero A1 or force = rate of change of momentum M1 force on trolleys equal and opposite A1 or impulse = change in momentum M1 impulse on each trolley is equal and opposite A1 (ii) 1. M1V1 = M2V2 B1 2. E = ½ M1V12 + ½ M2V22 B1 (iii) 1. 2 2 2 2 1 2 1 ()2 () 2 2 KE mv m mvm mv m p m = = = = 2. As trolley B has a smaller mass, it has a larger kinetic energy because momentum is the same/ constant for both trolleys. B1 M1 A0

3 3 (a) Phase difference = 2t T = 2(2.5−1.5) 3 = 2 3 = 120o A1 (b) At t = 0.75 ms, resultant displacement = sY + sZ = 4.0 – 1.0 = 3.0 µm A1 (c) I  A2 C1 Thus I = k (4)2 and IZ = k (2)2 = 0.25I A1 (d) v = f =  T C1 so  = vT = 330 (3.0 x 10−3) = 0.99 m A1 4 (a) (i) Current in R4 or R1 = 0.30 + 0.30 = 0.60 A C1 R = V / I M1 = 2.4 / 0.60 = 4.0 Ω A0 or p.d. across R3 or R2 = 2.4 / 2 = 1.2 V C1 R = V / I M1 = 1.2 / 0.30 = 4.0 Ω A0 (ii) E = 2.4 + 2.4 + 1.2 C1 = 6.0 V A1 or total resistance = 10 Ω C1 E = 10 × 0.60 = 6.0 V A1 (b) total resistance increases B1 current decreases (in battery) so total power decreases B1 5 (a) (i) It is to increase the magnetic flux linkage between the coils B1 Comments: Most candidates answered correctly. (ii) It is to reduce energy losses B1 by reducing induced currents B1 (b) (i) maximum VOUT = (Ns/ Np) Vp = (625 / 25 000) 12 000 = 300 V A1 (ii) r.m.s. current = peak current /√2 = 300 / (640 × √2) = 0.33 A A1

4 (iii) positive sinusoidal squared shape, with P = 0 at t = 20, 40 ms and maximum P shown as 140 W B1 two complete cycles, with period = 20 ms B1 (c) For sinusoidal squared graph, mean power is half the peak power (140 W) B1 6 (a) electron diffraction/ neutron diffraction B1 (b) ½mv2 = eV ½ × 9.11 × 10–31 × v2 = 1.60 × 10–19 × 4800 v = 4.1 × 107 m s–1 C1  = h / mv = 6.63 × 10–34 / (9.11 × 10–31 × 4.1 × 107) C1 = 1.8 × 10–11 m A1 (c) (i) Any two points, 1 mark each B2 • photons are (discrete) packets of energy • energy of photons depends on frequency (of EM radiation) • electrons can only absorb a single photon (of energy) • emission only possible if photon energy is at least the work function B1 (ii) work function = hf = 6.63 × 10–34 × 6.93 × 1014 C1 = 4.59 × 10–19 J = 4.59 × 10–19 / 1.60 × 10–19 = 2.87 eV A1 -200 -100 0 100 200 0 10 20 30 40 P / W t / ms

5 7 (a) Half-life of a radioactive nuclide is defined as the time taken for half of the original number of radioactive nuclides in a sample to decay on average. B1 (b) The beta decay is exothermic / releases energy. B1 Thus, the total binding energy of Y90 39 and 0 1− e is more than that of Sr 90 38 . B1 Since 0 1− e has no binding energy, Y90 39 has a greater binding energy. (c) (i) 1. A= N 3.7 x 106 = ln 2 (27.7)(365)(24)(60)(60)N C1 N = 4.66 x 1015 A1 2. 90 g of strontium contains 6.02 x 1023 atoms. Mass of 4.66 x 1015 atoms = 4.66×1015 6.02×1023 (90/1000) C1 = 6.97 x 10−10 kg A1 OR Mass of 4.66 × 1015 atoms = 𝑁(90𝑢) = (4.66 x 1015) (90) (1.66 x 10-27) C1 = 6.97 × 10−10 kg A1 Mass of 4.66 × 1015 atoms =𝑁(38 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 + 38 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑡𝑜𝑛𝑠 + 52 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠) = (4.66 x 1015) (38 x 9.11 x 10-31 + 38 x 1.67 x 10-27 + 52 x 1.67 x 10-27) = 7.01 × 10−10 kg (ii) A = Aoe−t A A0 =e−( ln 2 27.7)(5.0) C1 = 0.882 A1 8 (a) Energy has discrete/fixed values. B1 Comments: Generally p oorly answered. Use of the word packet does not connote quantised. B1

6 (b) (i) (ii) In the visible region, the intensity of the emitted radiation increases with increasing wavelength. B1 The red region has the longest wavelength and according to the graph highest intensity at 1100 K B1 The hot object is perceived by the eye as glowing red. A0 (c) (i) The wavelength , max, that corresponds to the peak intensity of the emitted radiation. The higher the temperature, the shorter the max. B1 (ii) advantage: The radiation can be detected at a distance. Hence there is no need for contact between device and the body to measure its temperature. B1 disadvantage: At lower temperature, the peak cannot be easily identified. B1 (d) (i) Electric force of attraction on electron provides the centripetal force for its motion about the nucleus. B1 e2 4or2 = mv2 r C1 v = √ e2 4omr = √ (1.6 x 10-19) 2 4(8.85 x 10-12)(9.11 x 10-31)r M1 = 15.9 √r A0 V visible region in the range from 400 nm to 700 nm. A1 I

7 (ii) v = nh 2mr = 15.9 √r n = 1 so h2 42m2r2 = (15.9)2 r C1 r = h2 42m2(15.9)2 = 0.053 nm A1 (iii) Energy = potential energy + kinetic energy = − e2 4or + 1 2mv2 = − (1.6 x 10-19) 2 4(8.85 x 10-12)(5.3 x 10-11) + 1 2 (9.11 x 10-31) (15.9)2 (5.3 x 10-11) C1 = − 2.17 x 10−18 J M1 = − 2.17 x 10−18 1.6 x 10−19 eV = − 13.6 eV A0 (e) (i) For minimum wavelength, n is infinity  = 1 1.097x 107 = 91.2 nm A1 (ii) E = hc 