NYGH-2014-S4-EOY-Physics-Answers

2014 Sec 4 End-Of-Year Examination Physics Marking Scheme Paper 1 Qn 1 2 3 4 5 6 7 8 9 10 Ans D C B D C D B B D C Qn 11 12 13 14 15 16 17 18 19 20 Ans D C B A A D A B B B Qn 21 22 23 24 25 26 27 28 29 30 Ans D C D C A A C B A B Paper 2 Section A 1 (a) upward is positive OR downward is negative (b) It moves with constant negative acceleration until it hits the ground at 0.60 s it rebounds upward with velocity 6.0 m s-1 and decelerates uniformly to 0.0 m s-1 when it reaches the highest point at 1.20 s (c) Total distance travelled = ½ (2.0 + 8.0) m s-1 × 0.6 s = 3.0 m (c) Displacement = ½ × 6.0 m s-1 × 0.6 s - 3.0 m = -1.2 m or 1.2 m downwards 2 (a) appropriate scale e.g. ( 1.0 cm rep 20 N) T 1 = 49  2.0 N T 2 = 85  2.0 N (b) Shift the two fixed pulleys M and N closer towards each other OR such that the cables are vertical 3 (a) Moment of a force about a point is the product of the force and the perpendicular distance of its line of action/the line of action of the force from the point. (no mark for “line of action” only, or use of other terms "force line", "line of force") (b) Apply Principle of Moments about the pivot, F (0.30 m) = 300 N (0.24 m) F = 240 N (c) The padding increases the contact surface and reduces the pressure on her legs 4 (a) time taken = 2 × 420 m / 1500 m s -1 = 0.56 s (b)(i) No, the water particles vibrate/oscillate in the same direction as/parallel to the direction of propagation of the ultrasound in water / travels as a longitudinal wave (b)(ii) 10 ms (c)(i) pressure due to water = (420 m × 1030 kg m -3 × 10 N kg-1) pressure = 100 kPa + (420 m × 1030 kg m -3 × 10 N kg-1) = 100 kPa + 4326 kPa = 4400 kPa (c)(ii) using ( 4426 kPa × 2.0 mm 3 ) = (100 kPa × v) v = 88 mm 3 T1 T2 98 N

2014 EOY S4 Physics Nanyang Girls’ High School Setters: TBH, MS Physics 5 (a) Inverted F (b) Image is real and inverted Correct image and object distances with a clear scale 2 light rays (with arrows) from one point of the object converging on the corresponding point on the image (c) Length must be correctly read off from ear lier diagram. Answers from calculations are not accepted. Answers with no clear indication of how the answer was obtained from the diagram are not accepted. Focal length = 15.0  5.0 cm (d) the image will be dimmer / fainte r / less bright / darker. 6 (a) f = v/  = 20 cm s -1/ 10.0 cm (b) v = f  = 2.0 Hz (2sf) = 2.0 Hz x 6.3 cm =12.6 cm s-1 7 (a) p.d. across X = 12.0 – 4.0 = 8.0 V I = V/R = 4.0/3.0 = 1.33 A Henc e, resistance of X = 8.0/1.33 = 6.0 Ω OR using potential divider method: (X +