2023 JPJC Prelim H2 Phy P2 Solutions

2023/JPJC/Prelim/9749/02 [Turn over Answers to 2023 JC2 Preliminary Examination Paper 2 (H2 Physics) Suggested Solutions: No. Solution Remarks 1(a)(i) 6 4 50 10 0.125 T4.00 10B A φ − − ×= = =× 5 0.01 0.102550 4.00 BA BA φ φ ∆∆∆=+=+ = ∆B = (0.125 T)(0.1025) = 0.01 T (1 s.f.)  B = (0.13 ± 0.01) T [1] value of B B ∆ [1] value of ∆B to 1 s.f. [1] answer 1(a)(ii) The percentage (or fractional) uncertainty of φ is more significant than that of A. [1] 1(b)(i) The perpendicular distance from A to line of action of T is given by d = 1.2 sin 50° Taking moments about A, (36)(0.45 cos 60°) = T (1.2 sin 50°) (36)(0.45cos60 ) 1.2sin50 8.8115 8.8 N (shown) T = = =   OR Anti-clockwise moment = Ty (1.2 cos 60°) + Tx (1.2 sin 60°) = (T cos 20°)(1.2 cos 60°) + (T sin 20°)(1.2 sin 60°) [1] moment equation [1] numerical answer before 8.8 N T d A 50°

2 2023/JPJC/Prelim/9749/02 No. Solution Remarks 1(b)(ii) Let Fx and Fy be the horizontal and vertical components of F respectively. Consider vertical forces: 36 N = T cos 20° + Fy  Fy = 36 − 8.81 cos 20° = 27.72 N Consider horizontal forces: Fx = Tx = T sin 20° = 8.81 sin 20° = 3.01 N 22 2 2 3.01 27.72xyF FF= += + = 27.9 N [1] value of F y [1] value of F x [1] answer 2(a) The phase difference is due to the extra distance x travelled by the wave from the student to the wall and back to him (x = 0.60 m) and also the phase change of π rad due to reflection at the wall. Phase difference 2 2 (0.60) 0.40 4 rad (or 0 or 2 rad) xπφπ λ π π ππ = + = + = [1] value of x [1] substitution [1] answer 2(b)(i) When x1 = 1.10 m, A1 = 1.2 × 10−8 m When x2 = 1.70 m, A2 = ? 821 1 21 12 2 8 1.101.2 10 1.70 0.78 10 m Ax x AAAx x − −  =→= =×   = × Assumption: no energy is absorbed by the wall upon reflection [1] substitution [1] answer [1] assumption 2(b)(ii) Since φ = 4π rad, they are in phase, so Amplitude of resultant wave = (1.2 + 0.78) × 10−8 m = 2.0 × 10−8 m [1] answer 2(c) 22 28 6 1 8 11 1 62 2.0 101.0 10 1.2 10 2.8 10 W m rr r r AA AA Ι ΙΙΙ − − − −−    ×= →= = ×   ×  = × [1] substitution [1] answer 3(a)(i) The electric field strength E at a point is the negative of the potential gradient dV dx− at that point. [1] 3(a)(ii) The equipotential lines are equally spaced with the same potential difference between consecutive lines. [1]

3 2023/JPJC/Prelim/9749/02 [Turn over No. Solution Remarks 3(a)(iii) Since VE d= , we have ( ) 2 600 600 2.5 10 E − −−= × 44.8 10E = × NC−1 [1] for correct substitution [1] for correct answer 3(b) [1] for six uniform field lines between the