2022 CWSS Prelim 4NA SCI (PHYSICS) MS for students

1 2022 Prelim 4NA Sci (Phy) Suggested Answers Paper 1 1 D 2 A 3 A 4 D 5 B 6 A 7 B 8 A 9 C 10 B 11 C 12 C 13 B 14 B 15 D 16 C 17 D 18 B 19 C 20 D Paper 2 Incorrect s.f. for answer(s) – minus one mark for whole paper Section A Qn Suggested Answer Mark Allocation Remarks 1(a) 1(b) 1(c) Electronic balance Micrometer screw gauge Stopwatch 1 correct – [1]; 2-3 correct – [2] (a) the word “balance” must be present (can accept “beam balance); “electronic” must be spelt correctly (b) can accept “Micrometer” 2(a) Moment = F x perpendicular d = 30 x (50+8) = 1740 N cm Direction of moment is anticlockwise [1] [1] [1] The perpendicular distance is from the pivot (the wheel) to the line of action of the 30N force. Direction of moment can never to left/right/up/down; only 2 possibilities – clockwise or anticlockwise. 2(b) Idea of “Extending the handle increases the perpendicular distance from the upward force to the pivot, thus creates a greater anticlockwise moment” in a logically sound sentence. [1]

2 3(a) (i) (ii) All data points accurately plotted - [1]; Smooth best-fit curve passing through all points – [1]; Each point must be marked with a cross, with the centre of cross falls accurately of the correct number. Best-fit line must be single and continuous, without obvious kinks and abrupt turns 3(a) (iii) Between 40 to 46 s [1] (e.c.f. from the graph) Do not read to more d.p. than possible from the graph (precision is half the smallest division) 3(b) Design A or B. With a logical reason of faster heat loss using either “greater total surface area exposed, increasing rate of radiation emission” Or “more space for more air flow” [1] 4(a) series [1] 4(b) R = V/I, 8 = 4.0 / I, I = 4.0 / 8 = 0.50 A (1,2,3 s.f.) [1] 4(c)(i) Total R = 8 + 1 / (1/4 + 1/4) = 10 Ω (1,2,3 s.f.) Parallel part: 1 / (1/4 + 1/4) OR Series part: 8 + parallel Final answer [1] [1] 4(c)(ii) The ammeter reading increases. [1] Adding another resistor in parallel provides another path for current to flow, increasing the total current.

3 Section B Qn Suggested Answer Mark Allocation Remarks 5(a)(i) Fres = ma, 0.32 = 0.40 x a, a = 0.32/0.40 = 0.80 m/s2 (2 s.f.) [1] Answer should be in 2 s.f. 5(a)(ii) Rate of change of velocity OR Change of velocity per unit time. [1] 5(a)(iii) Distance from second dot is obviously greater than distance between first and second dot [1] The trolley accelerates at 0.80 m/s2 (a positive number), meaning it is getting faster. 5(b)(i) a = gradient of s-t graph = (3.5-0) / (5.0-0) = 0.70 m/s2 (2s.f.) Reading graph and sub in gradient calculation: = (3.5-0) / (5.0-0) Final answer: = 0.70 m/s2 [1] [1] 5(b)(ii) Mentioning of any resistive force opposing the pulling force e.g. friction between trolley and track, pulling force due to ticker-tape timer and the tape etc. OR Ticker-tape timer is less accurate than the sensor (must be comparison between the two instruments) [1] 5(b)(iii) D = area under graph = ½ x 5 x 3.5 = 8.75 m (2 or 3 s.f.) [1] (e.c.f. same reading from (b)(i) ) 5(b)(iv) Save = D / t = 8.75 (b(iii)) / 5.0 = 1.75 m/s (2 or 3 s.f.) [1] (e.c.f. from (b)(iii) ) 6(a)(i) W = mg = 80 x 10 = 800 N [1]

4 6(a)(ii) Work = F x d = W x d = 800 (a(i)) x 8.4 = 6720 J (2 or 3 s.f.) OR (Energy at A) + W = (Energy at B); 0 + W = (GPE at B) W = GPE at B = mgh = 80 x 10 x 8.4 = 6720 J (2 or 3 s.f.) [1] (e.c.f. from (a)(i) ) 6(a)(iii) P = work / t = 6720 (a(ii)) / 16 = 420 W [1] (e.c.f. from (a)(ii)) 6(b)(i) Energy cannot be created or destroyed. It can only be changed from one form to another. The total energy of an isolated system remains constant. [1] 6(b)(ii) KE = loss of GPE = mgh = 80 x 10 x (8.4-2.6) = 4640 J Identify loss of GPE: = 80 x 10 x (8.4-2.6) Final answer: = 4640 J [1] [1] 6(c)(i) P = IV, 2200 = I x 240, I = 2200/240 = 9.2 or 9.17 A (2 or 3 s.f.) [1] 6(c)(ii) 10 A The fuse rating has to be slightly larger than 9.17A so that it does not melt when the lift operates normally but melts before overheating / prevent overheating. [1] 7(a) A wave that has its direction of travel parallel to the direction of particle vibration. [1] A wave travels parallel to the direction of vibration; NOT “a wave is parallel to …” 7(b)(i) 2d = s x t = 1500 x 0.14 = 210 m d = 210 / 2 = 105 m [1] [1] 7(b)(ii) In space there is vacuum / no matter / no particles present to transfer sound. OR Sound requires matter / particles / medium to transfer. [1]

5 7(c)(i) Amplitude lower than A (smaller than 1 division) [1] Upper and lower amplitude both smaller than 1 division 7(c)(ii) Trace C has a lower frequency than A, because sound in C has a lower pitch than sound in A. OR Trace C has a greater period than A OR Trace C has fewer number of completed waves during the same time. [1] 7(d)(i) 0.0120 m (precision = 0.0001 m) [1] 0.012 m is actually incorrect. When doing unit conversion, do not change the precision – the original s.f. 7(d)(ii) v = f λ, 1500 = f x 0.0120 (d(i)), f = 1500 / 0.0120 = 125 000 Hz [1] (e.c.f from (d)(i))