2025 ASRJC Prelim H2 P2 MS

1 9749/02/ASRJC/2025Prelim [Turn Over Anderson Serangoon Junior College 2025 H2 Physics P2 Prelim Mark Scheme Paper 2 (60 marks) E – Easy, A – Average, D – Difficult ECF Error carried forward SF Significant figures error M0 No A marks awarded AE Arithmetic error BOD Benefit of doubt ^ More is needed in answer POT Power of ten error CON Contradictory response XP Wrong physics TE Transcription error IR Irrelevant (part) response 1ai g = (4𝜋2 × 1.50) / (2.482) = 9.63 m s−2 Examiner’s comments: Most answers were correct. A small number of students incorrectly left the answer in 1 or 2 s.f. though the least s.f. of data in the question is 3. E A1 1aii percentage uncertainty = 2 + (3 × 2), OR fraction uncertainty = 0.02 + (0.03 × 2) actual uncertainty = 0.08 × 9.63 = 0.8 m s−2 Examiner’s comments: Many answers showed lack of understanding/familiarity of use of percentage uncertainty. A C1 A1 1bi ω = 2π / T = 2π / 4.0 = 1.57 rad s–1 v0 = ωx0 = 1.57 × 5.0 = 7.9 cm s–1 (2 s.f.) Examiner’s comments: Period and amplitude have to be read to half the smallest square. Some answers were obtained through the gradient approach, but since the tangent was not accurately drawn, their answers did not fall within the acceptable range. Hence, in this case the calculation approach is preferred. A B1 B1 1bii • initial pull was to the right/initial motion is toward the left • distance from X to trolley at equilibrium is 20 cm • initial motion undamped, then motion becomes (lightly) damped at/from 12 s • maximum speed at 1 s, 3 s, etc. / stationary at 2 s, 4 s, etc. Any two points, 1 mark each Examiner’s comments: Since this question refers to the given graph, quantities must be quoted where relevant (e.g. time when damping begins), except if the quantities had been determined in part (b)(i), as per instruction of the question. The key word “exponentially” is often missing when describing decrease in amplitude (e.g. amplitude decreases exponentially”. A B2

2 9749/02/ASRJC/2025 Prelim 1biii sketch: minimum L shown as 15 cm and maximum L shown as 25 cm minimum v shown as –7.9 cm s–1 and maximum v shown as +7.9 cm s–1 Examiner’s comments: Many did not draw the ellipse shape properly, which resulted in mark deduction. D B1 B1 2ai change in momentum = (1.40 – 0.40 – 0.80 sin 30) x 3.0 = 1.80 kg m s–1 Examiner’s comments: Unable to determine the net force acting on the block for the specified time interval. Working for net force and the concept of Ft was not clearly demonstrated to meet the demand of a “Show” question. Many did not read graph to half smallest division. Need to indicate the smallest division even if the last digit is a “zero”. A M2 A0 2aii resultant for