2023 JPJC Prelim H2 Phy P4 Solutions

2023/JPJC/Prelim/9749/04 [Turn over Answers to 2023 JC2 Preliminary Examination Paper 4 (H2 Physics) Suggested Solutions No. Solution Remark 1(a) x = 15.0 15.0 15.0 cm2 + = Period T = 20.8 20.7 1.04 s2 20 + =× [1] - for correct measurements with units - 1 d.p in cm - repeat [1] - 1 or 2 d.p in timing - repeat - T in 3 s.f. or 4 s.f. depending on the d.p. of t1 and t2 t ≥ 20.0 s 1(b) x/cm N Time for N oscillation Period T/s T4/s4 t1/s t2/s 11.0 25 24.2 24.1 0.966 0.871 13.0 25 25.1 25.2 1.01 1.04 15.0 20 20.8 20.7 1.04 1.17 17.0 20 21.4 21.6 1.08 1.36 19.0 20 22.1 22.2 1.11 1.52 21.0 20 22.9 22.8 1.14 1.69 [1] - headings and units - 6 sets of data [1] - d.p. of raw data - t ≥ 20.0 s [1] s.f. of processed data [1] correct calculation, allow 1 slip Don’t accept x = 0 cm. This will not be considered one set of data. 1(c) Refer to attached graph. [1] axes: units, scale [1] plotted points accurate to half of smallest division [1] best fit line 1(c) Given T4 = Px + Q Graph of T4 vs x is plotted, where P is the gradient and Q is the y-intercept. [1] - Big triangle - substitution of gradient coordinates - linearisation

2 2023/JPJC/Prelim/9749/04 ( ) ( )( ) 41 4 1.67 0.89Gradient= 0.0830 20.8 11.4 = 0.0830 s cm Substitute 20.8,1.67 into the equation, 1.67 0.0830 20.8 0.0564 s P Q Q − − =− = + =− [1] P calculated correctly with units [1] Q calculated correctly with units

3 2023/JPJC/Prelim/9749/04 [Turn over

4 2023/JPJC/Prelim/9749/04 No. Solution Remarks 2(a)(i) L0 = 7.5 +7.5 7.5 cm2 = [1] correct measurements with unit and d.p 2(a)(ii) Volume V = (1.9×10−3)(1.9×10-3)(2×7.5×10−2) = 5.42×10−7 m3 [1] ans 2(b)(i) L = 8.8 8.8 8.8 cm2 + = Extension e = 8.8 – 7.5 = 1.3 cm = 0.013 m Force F = 100×10−3 × 9.81 = 0.981 N [1] - both e and F calculated correctly - repeat measurement for L - answer for F in 2 or 3 sig. fig 2(b)(ii) m/kg L/m e/m F/N 0.000 0.075 0.000 0.000 0.100 0.088 0.013 0.981 0.200 0.116 0.041 1.96 0.300 0.165 0.090 2.94 0.400 0.212 0.137 3.92 0.500 0.261 0.186 4.91 [1] - headings and units - 6 sets of data (award full credit if m = 0.000 kg not included in the table) [1] - d.p. of raw data - m in 3 d.p - s.f of processed data [1] correct calculation, allow 1 slip 2(b)(iii) Refer to attached graph. [1] - plotted points accurate to half of smallest division - best fit curve / line 2(b)(iv) When the extended length is 2L0, the extension e is L0 = 0.075 m and force F is 2.7 N. Energy stored = area under the graph 1 (0.981 1.96)(0.028)(0.013)(0.981)22 += + (1.96 2.7)(0.034) 2 ++ = 0.127 J [1] correct calculation No marks awarded if best fit curve / line does not pass through origin 2(b)(v) Energy stored per unit volume = 7 0.127 5.41 10 −× = 2.35 ×105 J m−3 [1] correct calcu