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 approaches in determining whether the collision was elastic. Since total KE is not conserved, the collision is not elastic. OR Since relative speed of approach is not equal to the relative speed of separation, the collision is not elastic.

Anglo-Chinese Junior College 2022 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2022 Page 3 of 14 Q2 (a) Electric potential at a point in an electric field is defined as the work done per unit positive charge by an external agent in bringing a small test charge from infinity to that point, Most students only obtained 1 out of 2 marks. Wrong terms used include “work done” instead of “work done per unit charge”, and “stationary charge” even though the charge is being moved. Essential terms that were left out include “external agent” and “positive”. without producing any acceleration. This was left out by some students. (b) 2 2 0 050 0 200 100 V 1 400  = = . . r V V V This part was well answered. 2 2 2 -1 2 500 N C 1 0 050 8000 0 200  = =    r E. . E E This part was well answered. (c)(i)1 Refer to sketches − R < r < R : Horizontal line Vinside = + 400 V This part was well answered. • (r < − R) and (r > R) : Curve V = + o V0.050m = + 400 V V0.200m = + 100 V This part was well answered. A few students did not sketch the graph to pass through the points at 0.1 m and 0.2 m. Some answers did not include range from -0.2 m to 0.2 m. (c)(i)2 0 < r < R : Horizontal line Einside = 0 This part was well answered. • (− R < r < 0) and (R > r > 0) : Curve & direction o (When r = +, E= +) (When r = −, E = −) o E0.050m =  8000 N C-1 E0.200m =  500 N C-1 Wrong answers included taking the wrong direction of E as positive, not sketching the graph to pass through the points at 0.1m and 0.2m. Some answers did not include range from -0.2m to 0.2 m.

Anglo-Chinese Junior College 2022 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2022 Page 4 of 14 (c)(ii)1 • − R < r < R : Vinside = + 300 V (horizontal line) • (r < − R) and (r > R) : Curve V = + (r < − R) and (r > R) : horizontal line at V0.200m = 0 This was very poorly answered even for students who had full marks for (c)(i). Students need to consider the graph for sphere A, and then determine the scalar sum of graphs for sphere A and sphere B. (c)(ii)2 • Between E0.050m and E0.200m : Curve & direction o (r = +, E= +) (r = −, E = −) o E0.050m =  8000 N C-1 o E0.200m =  500 N C-1 (r = +, E= +) (r = −, E = −) • Einside = 0 Ebeyond 0.200m = 0 This was very poorly answered even for students who had full marks for (c)(i) where some of the graphs drawn clearly did not apply the equations for V and E. Students need to consider the graph for sphere B, and then determine the vector sum of graphs for sphere A and sphere B.

Anglo-Chinese Junior College 2022 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2022 Page 5 of 14 0 0 r / m V / V − 0.20 − 0.10 0.10 0.20 0 0 0 0 r / m E / N C-1 0.10 0.20 − 0.20 − 0.10 500 − 500 0 A B 400 8000 300 100

Anglo-Chinese Junior College 2022 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2022 Page 6 of 14 3 (a) First Law of Thermodynamics states that the increase in internal energy of a system A number of students incorrectly stated “change” in their answers. is the sum of the heat supplied to the system and the work done on the system. (b) 6(2.26 10 )(5.0)toq = 5.0 5.0()0.598 1000V = − Some students are still not finding change of quantities the right way. 5 5.00 5.00(1.01 10 )( )0.598 1000 onw =−  − A number of students found work done by the system instead. 71.05 10 J (3 s.f.) to onU q w = + = (c)(i) As the product of p and V at state C is greater than the product of p and V at state D, OR to onU q w = + Since qto is 0 and won is negative due to expansion, This part was well answered. State C is at a higher temperature. (c)(ii) section of cycle heat supplied to ga