Title 2022 GCE A level H2 Physics 9749 Ans.pdf ✅

2022 H2 Physics Solution 1 Paper 1 1 Ans: A Given that X and Y are initially opposite, then X – Y = (+ve) – (-ve) = larger in magnitude. After Y rotates 180°, then X – Y = (+ve) – (+ve) = smaller in magnitude. 2 Ans: C sx = uxt 100 = (u cos θ)(2.0) u cos θ = 50 sy = uyt + 1⁄2 ayt 2 0 = (u sin θ)(2.0) + 1⁄2 (-9.81)(2.02 ) u sin θ = 9.81 tan θ = 9.81 / 50 = 11° 3 Ans: B Using PCOM, mPuP + mQuQ = mPvP + mQvQ mp(2.0) + 0 = mp(-0.50) + mQvQ (2.5)mp = mQvQ Since the collision is elastic, vQ – vP = uP – uQ vQ – (-0.50) = 2.0 – 0 vQ = 1.5 Therefore, mP mQ = 1.5 2.5 = 3 5 4 Ans: B T + W = U T = (0.5V)(1030)(9.81) – (200)(9.81) = (0.5)(4 3 π)(0.5003 )(1030)(9.81) – (200)(9.81) = 680 N

2022 H2 Physics Solution 3 11 Ans: A Average KE is proportional to temperature in Kelvin. = 160 + 273.15 80 + 273.15 3502 = 433.15 353.15 = 390 m s-1 12 Ans: C At constant temperature, total KE remains constant. Therefore final KE = initial KE = 3 2 nRT = 3 2 pV = 3 2 (1.0 × 105 )(0.010) = 1500 J 13 Ans: B Q = 1 2 ( 1 2 mv2 ) mc∆T = 1 4 mv2 ∆T = v 2 4c 14 Ans: C x = x0 sin ωt x = 0.30 sin 2π 5.0t v = dx dt = 0.38 cos 1.3t 15 Ans: A All B, C, and D are examples of resonance. 16 Ans: D v = fλ = λ T = 0.8 0.2 = 4.0 m s-1 17 Ans: B Period = 6 squares on x-axis × 0.050 ms = 0.30 × 10-3 s Frequency = 1/period = 3.3 kHz Since X and Y are positions that give max amplitude, 5 cm = half a wavelength Hence, Wavelength = 10 cm 18 Ans: B Using x = λD a , by trial and error, we get B as the answer. For option A: x = (700 nm)(15) 4.0 mm For option B: x = (20 mm)(15) 50 mm For option C: x = (450 nm)(15) 2.0 mm For option D: x = (10 mm)(15) 200 mm
2022 H2 Physics Solution 4 19 Ans: A time taken to travel through the plates horizontally, t = y v deflection x = 0 + 1 2 (a)(t2 ) = (1 2 )(qE m )(y 2 v 2) = (1 2 )(eV md)(y 2 v 2) = eVy2 2mdv2 20 Ans: A Current is constant throughout. Using I = nAve, I = n( πd 2 4 )(v)(e) v = K d 2, where K = 4I nπe 21 Ans: D Let resistance of P and voltmeter be R. Let resistance of Q be RQ. Effective resistance of P // voltmeter = 0.5 R Using potential divider method, 0.5R 0.5R + RQ × 9.0 = 6.0 4.5R = 3.0R + 6.0RQ R RQ = 4.0 22 Ans: B e.m.f. = p.d. across 65 cm the resistance wire = 0.65 × 14.3 = 9.3 V 23 Ans: C Like current attracts, so the forces should be attractive. These forces are action-reaction pair, so they should be equal in magnitude. To derive, you need to apply both RHGR to get the B due to each current, and FLHR on each wire.. 24 Ans: D When θ = 0°, F will be maximum. As wire rotates, F = F0 sin θ 25 Ans: B Flux = (perpendicular B)(area) = (65 × 10-6 )(sin 60°)(12 × 10-4 ) = 6.8 × 10-8 Wb