ACJC 2022 Prelim P3 Guide

Anglo-Chinese Junior College 2022 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2022 Page 1 of 14 Annotations used in marking BOD - Benefit of doubt ECF - Error carried forward POT - Powers of ten error TE - Transfer error CE - Calculation error XP - Wrong physics ENG - Generally bad english, phrasing and expression PP - Poor presentation of answers Q1 Suggested Answer Marker’s Report (a)(i) This part was generally well done. Some students omitted one of the forces and were unable to get full credit. Forces correctly labelled (ruler should be used) Correct direction indicated (T longer in length than component of Wc along slope) (a)(ii) Taking mass B as the system -2 (8.3 9.81) 54 8.3 3.30 m s BBm g T m a a a −=  − = = This question was answered well by students who understood the concepts.These students obtained full credit. The rest attempted to calculate based on the entire system but failed due to the lack of data. Others were confused with the resolution of the forces. Taking mass A as the system sin50 54 ( 9.81sin50 ) 3.30 AA AA T m g m a mm − = −   =  4.99 kgAm = 5.0 kg (2 s.f.) (shown)= (b)(i) 3 Force on block A gradient 2.00 ( 10.00) 3.00 10 2670 N (3 s.f.) − = − − −=  = Most errors on this part are due to the wrong exponent. Force on block A Force on block C 2670 N =− =− Most students did not account for N3L in this part and hence failed to obtain full credit.

Anglo-Chinese Junior College 2022 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2022 Page 2 of 14 (b)(ii) 2.00 5.0 A A A pv m= −= The vast majority of students were successful in this part. -10.40 m s=− A number of students forgot to express the answer in appropriate s.f. (b)(iii) Total momentum conserved: 10.00 ( 2.00) 8.00 N s if i fA fC f C i f A pp p p p p p p = =+ = − =− − − =− Students who understood the concepts performed well in this question, obtaining full credit. -18.00 0.80 m s10 Cv =− =− Students who faltered in this question, usually were confused with the directions of the momentum. Some students who used the SUVAT approach to obtain velocity values were successful. 2 2 22 22 1Total KE before collision 2 1 10.00(5.0)( )2 5.0 10 J 1Total KE after collision [ ]2 1[(5.0)( 0.40) (10)( 0.80) ]2 3.6 J A A A C C mu m v m v = −= = =+ = − + − = OR -1 -1 relative speed of approach 2.0 m s relative speed of separation 0.80 0.40 0.40 m s AC CA uu vv =− = =− =− = Most students were familiar with the conservation of KE/relative speed approa