2021 Prelim H2Phy P2 Ans

1 2021 DHS H2 Physics Prelim Paper 2 Suggested Solutions 1 (a) time B2 (electric) current allow amount of substance allow luminous intensity any two of the above quantities, 1 mark each (b) (i) = 0.562 s A1 percentage uncertainty = (2% + 8%) / 2 (= 5%) C1 or fractional uncertainty = (0.02+0.08) / 2 (= 0.05) ΔT = 0.562 x 0.05 = 0.028 s C1 T = (0.56 ± 0.03) s A1 (ii) 1. total energy = max kinetic energy = ½ (200 x 10-3)(2 /T)2(5.00 x10-2)2 M1 = A1 2. total energy = = 3.15 x 10-2 J A1

2 2 (a) sum / total momentum of bodies is constant B1 or sum / total momentum of bodies before = sum / total momentum of bodies after for an isolated / closed system / no (resultant) external force B1 (b) (i) 4.0 vA = 6.0 vB C1 EK = ½mv2 C1 ratio = M1 = 1.5 A0 (ii) 0.48 = EK of A + EK of B = EK of A + (EK of A / 1.5) = 5/3 × EK of A C1 EK of A = 0.29 (0.288) J A1 (iii) curve starts from origin and has decreasing gradient M1 final gradient of graph line is zero A1 3 (a) (i) EK = ½mv2 = 0.5 × 0.40 × 0.302 = 1.8 × 10–2 J A1 (ii) (change in) kinetic energy = work done on spring / (change in) elastic potential energy C1 1.8 × 10–2 = ½× F × 0.080 C1 FMAX = 0.45 N A1 (iii) a = F / m = 0.45 / 0.40 = 1.1 m s–2 A1 (iv) 1. constant velocity / resultant force is zero, so in equilibrium B1 2. decelerating / resultant force is not zero, so not in equilibrium B1 (b) curved line from the origin M1 with decreasing gradient A1

3 4 (a) two sine waves in antiphase (one dotted, one solid line) B1 4 antinodes and 5 nodes shown B1 (b) a second sine wave of same wavelength on same axis B1 separated by a quarter of the wavelength B1 (c) plane parallel wavefronts before the opening B1 circular wavefronts showing diffraction after the opening with same wavelength and B1 greater than the gap width B1 5 (a) M1 M1 A0 (b) (i) From (a)   C1 A1 (ii) C1 A1

4 (iii) C1 A1 (c) Over-estimate. 0.20 V is actually the p.d. across the internal resistance cell r as well as that of the ammeter. B1 B1 6 (a) (i) to increase flux linkage (with secondary coil) B1 (ii) e.m.f. (induced only) when flux (in core/coil) is changing B1 constant / direct voltage gives constant flux / field B1 (b) (i) NS / NP = VS / VP C1 NS = (52 / 150) × 1200 = 416 turns A1 (ii) 0 ms or 7.5 ms or 15.0 ms or 22.5 ms A1 (c) (i) either mean power = V2/ 2R and V = 52 (V) C1 R = 522/ (2×1.2) = 1100 (1127) Ω A1 or mean power = V2/ R and V = 52 /√ 2 (= 36.8 V) C1 R = 36.82/1.2 = 1100 Ω A1 (ii) sinusoidal shape with troughs at zero power B1 only 3 ‘cycles’ B1 each ‘cycle’ is 2.4 W high and zero power at correct times B1 7 (a) In the magnetic field, the magnetic force acting on the ion

5 provides the centripetal force for the ion to move in uniform circular motion. Thus Since , which is independent of r. Assume that the time taken by the ion to cross the gaps