Raffles Institution FE 2019 ANS

2019 Y4 Phy FE Mark Scheme (MARKERS TO UPDATE MARKS/ANSWER/COMMENTS) Section A Multiple Choice Questions Qn Ans Qn Ans Qn Ans 1 C 11 A 21 B 2 C 12 A 22 C 3 C 13 D 23 A 4 D 14 C 24 B 5 B 15 C 25 D 6 B 16 B 26 C 7 D 17 B 27 A 8 A 18 C 28 A 9 D 19 A 29 B 10 D 20 B 30 B Section B Structured Questions 31 2 1 11 From 6 , 6 kg.m sUnit of (correct units for each quantity )m.m s kg.m .s R rv R rv     − − −− = = = = 32 (a) Acceleration is constant. Motion is along a (straight) line or all vectors are along a line. (b) (i) sbag = ubag t + ½ abag t2 = (−10.0)(4.00) + ½ (9.81)(4.00)2 = 38.5 m Note: allow both  answers (ii) sballoon = uballoon t + ½ aballoon t2 = (10.0)(4.00) + 0 = 40.0 m Separation between balloon and the sand bag = 40.0 + 38.5 = 78.5 m (c) (i) Magnitude =  (3.02 + 10.02) = 10.4 m s−1 Angle = tan−1 (10.0/3.0) = 73.3 above the 3.0 m s−1 velocity vector. (Reject angle w/o reference line, and reject reference to NSEW/bearings.) Note: As question states “calculate”, max [1] for scale drawing. (ii) Both sand -bags take the same time to reach the ground (because the horizontal wind has no effect on the vertical motion). 33 (a) Taking moments about support A, since the rod is not rotating,

anti-clockwise moments = clockwise moments FB  7.00 = 4.00  5.00 FB = 20.0/7.00 = 2.86 N (3 sf) (b) From Newton’s 2nd Law, since the rod is not moving, (reject other methods) total upward forces = total downward forces FA + FB = W FA = 5.00 − 2.86 = 2.14 N (allow ecf) 34 (a) loss in KE = ½mv12 – ½mv22 = ½ (2.79 x 105)(35.02 – 15.02) = 1.40 x 108 J (b) B [A1] – highest SHC, so increase in temperature is minimised [M1] OR C [A1] – lowest density, so lightest [M1] D [A1] – highest melting point, so least likely to melt [M1] (c) Initial acceleration - Distance covered = 640 m, time taken = 32 s Deceleration before region - Distance covered = 724.375 m, time taken = 23.75 s Travel at Constant speed in first region – Distance covered = 1483.625 m, time taken = 37.1 s Travel in second (safety) region – Distance covered = 1.00 km, time taken = 47.6 s Fastest speed in final region can only reach 36 m s-1 – Distance covered during acceleration = 342 m Time taken during acceleration = 12.0 s Distance covered during deceleration = 810 m Time taken during deceleration = 45.0 s Shortest time = 197 s (3sf) 35 When a neutral body is placed closed to a positively charged body, electrons will be attracted to the side facing the charged body. The far side becomes positively charged. The force of attraction is stronger than the force of repulsion. Note: positive charges cannot move. 36 (a) Vout = 8000 600+8000 𝑥 12.0

= 11.2 𝑉 (b) (i) 12.0 − 8.0 = 4.0V (ii) Apply R1/R2 = V1/V2 Resistance = 4000 Ω (c) Cooler, fan, aircon. (Reject fire alarm.) The de