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. Many students failed to note that loss in KE = gain in GPE, and gain GPE is proportional to height, hence the graph is a straight line. A A1 1c With a large splash, a large part of the kinetic energy of the child is transferred to the kinetic energy of the splashed water, as well as sound energy. This leaves the child with a small kinetic energy to do work against the pool water’s resistive force when he enters the pool. Hence, the child will be slowed down in a shorter distance when it enters the pool with a large splash. Examiner’s comments: This part was poorly done. Some students mentioned that large splash causes larger resistive forces from water. The majority who mentioned energy did not refer to the energy forms that were involved in the two cases. There were many vague descriptions that more energy was been lost by the child when it made a splash, did not mention the KE of the child to the KE or GPE of the water. Many students did not mention when the child enters the pool, work is done against water’s resistive force. D M1 M1 A0 2ai Total initial momentum = muA + muB = m (500 + 0) = 500m Examiner’s comments: This part was well done by most candidates. However, some candidates were penalized when they used terms such as mA (instead of using m) E A1 2aii1 mvA cos 60° + mvB cos 30° = mvA (0.50) + mvB (0.87) Examiner’s comments: This part was well done by most candidates. However, some candidates were penalized when they did not include the mass in their expression. A A1 h height kinetic energy 0 0

3 9749/02/ASRJC/2023PROMO [Turn Over 2aii2 (Taking upwards as positive) mvA sin 60°− mvB sin 30° = mvA (0.87) − mvB (0.50) Or (Taking downwards as positive) mvB sin 30° − mvA sin 60° = mvB (0.50) – mvA (0.87) Or mvA sin 60°+ mvB sin 30° = mvA (0.87) + mvB (0.50) Examiner’s comments: This part was well done by most candidates. However, some candidates were penalized when they did not include the mass in their expression. A A1 2aiii By Conservation of Linear Momentum, 500m = mvA (0.50) + mvB (0.87) 500 = vA (0.50) + vB (0.87) and 0 = mvA (0.87) − mvB (0.50) vA (0.87) = vB (0.50) Solving the two equations for vA, vA = 250 m s−1 Examiner’s comments: This part was poorly done as many candidates did not know how to make use of the momentums in the 2 directions. Common mistakes include equating 500m to the sum of the final momentums in the x and y directions. Candidates should realize that vectors in different directions cannot be added up without considering their directions. D M1 M1 A0 2b Δv = vf – vi = vf + (–vi) 22Δ 2 cos 60f i f iv v v v v= + −  ( )( ) 22Δ 250 500 2 250 500 cos 60v = + −  = 433 ≈ 430 m s−1 22 Δ 2 Δ cos θi f fv v v v v= + − ( )( ) 22500 250 433 2 250 433 cos θ= + − θ = 90° Or use sine rule sin sin60 500 430 90   = = A M1 A1 A1 vf −vi Δv 60° θ

4 9749/02/ASRJC/2023PROMO Examiner’s comments: Despite being tested on the same concept on other occasions, many candidates are still unsure of how to determine the change in vectors (when the vectors are in different directions). Also, candidates should note that they should not include the negative sign when applying the cosine rule. Others were penalized when they made assumptions in their working (e.g. assuming that the vector triangle is a right -angled triangle without proving it). Candidates should draw clear and well -labelled vector diagrams. Similar to the earlier part, candidates should realize that vectors in different directions cannot be added up/subtracted without considering their directions. 3a Moment = 0.30 x 0.29 cos 40° = 0.067 N Examiner’s comments: Generally well done. Students who are unable to get full credit usually had difficulties manipulating the angles and resolv ing vectors properly despite it being a rather straightforward scenario. They are encouraged to revise and practice how to resolve vectors correctly as this is a critical skill in the study of Physics. A small number of students did not get full credit due to rounding error. E1 – resolve perpendicular distance wrongly, i.e. using 0.29 sin 40° E2 - resolve weight wrongly, e.g. F = mg / sin 50° E3 – manipulated angles wrongly, e.g. sum of angles = 100° for right angle E C1 A1 3bi Volume of sphere = ( ) 34 0.04803π Upthrust = ( ) 341000 9.81 0.0480 0.26 1.18163ρgV =      = π ≈ 1.18 N Examiner’s comments: Generally well done. Students need to show all steps clearly, especially substitutions and pre-rounded value to score full credit. E1 – did not show pre-rounded value of 1.186 N E2 – wrong formula for volume of sphere A M1 A0 3bii Taki