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 force (on block) is zero (so) velocity is constant Examiner’s comments: Those who were unable to determine the net force in earlier part often encountered difficulty in this part. Some concluded the velocity wrongly with the correct net force. D B1 B1 2aiii 0 to 3.0 s: upward sloping straight line from the origin. 3.0 to 6.0s: horizontal line at non -zero value of momentum with no ‘step change’ in momentum at 3.0 s A B1 B1 x

3 9749/02/ASRJC/2025Prelim [Turn Over Examiner’s comments: Many could obtain the correct answer for 0 to 3 s only. Common mistakes seen for 3 to 6 s included a step changed at 3 s and a non-linear line drawn. 2bi total initial momentum is not zero (in the absence of an external force,) the total momentum can never be zero, hence stopping at different times. or force on each nucleus is equal in magnitude (by N3L) with different mass, they have different magnitudes of deceleration and hence stopping at different times. Examiner’s comments: Many students merely define COLM or stated “by COLM” without applying COM into the context of the question. Students need to state explicitly that the total initial momentum value is non-zero and to recognize that total final cannot be zero. By stating “total final momentum must be non-zero” merely regurgitated COLM and did not contextualize to the event when the nuclei stop at the same instant. Many students were not sensitive/clear with their choice of words used in explanations which could bring about different understanding, eg. “initial momentum is zero” is not the same as “total initial momentum is zero”. D M1 A1 A0 (M1) (A1) (A0) 2bii curve with positive final speed same change in momentum or total momentum remains constant or same force (N3L) since mass of P > mass of Q, change in velocity of Q > change in velocity of P or curve with positive final speed (Speed of approach = speed of separation) Speed of approach or speed of separation is positive value Hence, speed of Q > speed of P D B1 M1 A1 (B1) (M1) (A1)

4 9749/02/ASRJC/2025 Prelim Examiner’s comments: Many students merely stated COLM or RSS = RSA without applying them into the context of the question. 3ai (tangent to) line gives direction of force on a (small test) mass E B1 3aii (tangent to) line gives direction of force on a (small test) positive charge Examiner’s comments: This recall-type question was generally not answered well. E B1 3b similarity: lines are radial / greater separation of lines with increased distance from the sphere/ lines normal to the surface difference: gravitational lines directed towards sphere and electric lines directed away from sphere Examiner’s comments: The answer must take into account relevance to the “positively charged sphere” context. Some answers showed that candidates lack of use of proper terms such as “radial”, or incorrectly use the term “uniform”. A B1 B1 3ci E = Q / 4πε0r2 or E = kQ / r2 with k defined / substituted in 4.1 × 10–5 = [Q / (4π × 8.85 ×10–12 × 0.0252)] – [Q / (4π × 8.85 × 10–12 × 0.0752)] Q = 3.2 × 10–18 C Examiner’s comments: This question was mostly answered well. A C1 C1 A1

5 9749/02/ASRJC/2025Prelim [Turn Over 3cii smooth curve with gradient decreasing starting at (0, 4.1 × 10–5) to d-axis at (2.5, 0) smooth curve with gradient increasing from (2.5, 0) ending at (5, – 4.1 × 10–5) Examiner’s comments: The start and end of curve must be plotted to half the smallest square. A B1 B1 3ciii acceleration decreases (to zero at mid-point) then acceleration increases in the opposite direction / increasing negative acceleration Examiner’s comments: Many answers used as the verb “accelerated” which resulted in inaccurate descriptions. It is better to use the noun “acceleration” accompanied by its increase/decrease and its direction. (Note: deceleration is not always negative acceleration , as it depends on positive direction.) D B1 B1 4ai1 4ai2 total volume of molecules negligible compared to that of containing vessel molecules in random (and rapid) motion time of collision small compared with the time between collisions large number of (identical) molecules (any two of above, 1 mark each) Examiner’s Comments: Many students did not read the question carefully and stated assumptions that are already implied in the question such as elastic collisions and no intermolecular forces between molecules. E B1 B1 x x x

6 9749/02/ASRJC/2025 Prelim Students to take note that for assumption on comparing the volume of the molecules with the volume of the container, it must be the total volume of the molecules, not the volume of