2023 HCI Prelim P2 AS

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 1 2023 Preliminary Examination H1 Physics Paper 2 Suggested Solution Qn 1 Answer Marks (a)(i) Actual dimensions of 330 ml soft drink can radius r = 3.2 cm, height h= 12.2 cm (the volume of the can is greater than the 330 ml of soft drink contained) Acceptable range 3 5 cm 2 5 cm. r .   , 14 cm 10 cm h Volume of a can, 2V r h  Acceptable range for volume of can,    3550 196 cmV Accept if student quotes 330 cm3 A1 (a) (ii) Consider cans as being blocks of dimensions 2r x 2r x h. These blocks are stacked in the crate ……………………………. ………………………………. (independent of can dimensions) M1 M1 A1 3 1No of blocks in 1 m , 2 2N r r h     2 Empty space per block, 2 2 4 empty block canV V V r r h r r h r h                  2 2 3 Total empty space in crate 1 44 4 14 0 21 m emptyN V r hr h .          

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 2 (b)(i) M1 A1 (b)(ii) time for car to move 2 km = t Using 21 2s ut at  1 9 20002 21.1 s t t t     time for jet to move 2 km = t’ time difference = 1.2 s [A1] Alternatively, time for car to move 2 km = t   2 2 1 2 12000 9 2 21 1 s s ut at t t .     [M1] time for jet to move 2 km = t’ s1 distance travelled by jet in first 10 s s2 distance travelled by jet from 10 s to finish           2 1 2 2 1 2 2 10 10 5 10 250 m2 150 10 15 10 2 2000 1250 50 10 15 10 20002 22 3 s s s t t s s t t t .                         time difference = 1.2 s [A1] M1 M1 A1 Max Marks 9  100 1000 0 93600 3.1 s v u at t t               2 1 110 50 50 50 15 10 10 20002 2 1250 100 15 150 10 20002 15 200 3000 0 22.3 s t t t t t t t                      t t

2022 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 3 Qn 2 Answer Marks (a) Weight component of the block normal to the slope = 1.6 (9.81) cos 30 = 13.59 N In the normal direction, net force = 0 Normal contact force by the scale on the block = weight component of the block normal to the slope = 13.59 N By Newton’s 3rd Law of Motion, Normal contact force by the block on the scale (i.e. this force determine the reading on the scale) is equal and opposite to the normal contact force by the scale on the block. Thus, the reading on the scale = 13.59 / 9.81 = 1.39 or 1.4 kg. B1 B1 B1 A1 (b)(i) Consider the block: Let the velocity of the block immediately after collision be v. By conservation of energy, Loss in KE = Gain in GPE 2 -1 1 2 2 0.5