2023 ASRJC H2 physics Promo P2 MS_student

1 9749/02/ASRJC/2023PROMO [Turn Over Anderson Serangoon Junior College 2023 H2 Physics Promo P2 Mark Scheme Paper 2 (80 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 Table of Specifications (Paper 2) Question 1 2 3 4 5 6 7 8 9 Total % Easy 2 1 2 0 2 4 0 1 3 15 18.75 Average 5 5 5 10 5 6 5 7 5 53 66.25 Difficult 3 2 2 2 3 0 0 0 0 12 15.00 Marks 10 8 9 12 10 10 5 8 8 80 100.0 1a Work done (against frictional force) =loss in GPE – gain in KE 52 × (distance moved) = 330×(4.0 – 1.1) – 540 distance moved = 8.019 = 8.0 m Examiner’s comments: Most students get this correct. Some students applied Conservation of Energy wrongly, e.g loss in GPE = gain in KE + friction+ work done. Students should also note that there is gain in GPE when the child moves from the lowest point to Y and the loss in GPE is mg(4.0 -1.1). A C1 A1 1bi Ek = ½mv2 540 = ½ × (330/9.81) × v2 v = 5.67 = 5.7 m s-1 Examiner’s comments: Generally well done except some students made careless mistakes in substitution, using wrong value (330) for m. E A1 1bii Speed at point Z = horizontal component of velocity at point Y = 5.67 × cos 41o = 4.279 = 4.3 m s−1 Examiner’s comments: Some students did not show pre-rounded value of 4.279 m s−1. A M1 A0 1biii At point P, horizontal velocity, vx, is constant at 4.3 m s-1 tan 48𝑜 = 𝑣𝑦 4.3 vy = 4.776 m s-1 = 4.78 m s-1 Taking downwards as positive, 𝑣𝑦2 = 𝑢𝑦2 + 2𝑎𝑦𝑠𝑦 (4.78)2 = (−5.67𝑠𝑖𝑛41𝑜)2 + 2 × 9.81 × 𝑠𝑦 sy = 0.459 = 0.46 m D C1 C1 A1

2 9749/02/ASRJC/2023PROMO Examiner’s comments: Most students found this question challenging. Many students using v2 = u2+2as and substituted the incorrect initial and final velocities (swopped speed and vertical velocity). Students failed to apply the common approach to solving projectile motion problem (considering the motion separately in x and y directions and show clear workings using symbols and subscripts). Some students applied Conservation of Energy but calculated the height wrongly (either from Y to Z or from Z to P). 1biv Line with a negative gradient starting from a non-zero value kinetic energy when the vertical height is zero, and the straight line ends at a non-zero value of kinetic energy when the vertical height is h. Examiner’s comments: This part was poorly done. M