NYGH 2019-S3EOY-Physics P1 and P2 Ans

Confidential Markscheme for 2019 Sec 3 EOY PHYSICS Paper 1 1 . .5 6 .10 11 ..15 16 ..20 21.. 25 26.. 30 CADAC BDCCD ADBBB ADCDA ADDBB AACDD 3 D Accuracy is how close the values are to the true value, while precision is how close the values are to each other. A large number of students chose the option ‘A’, however the values are neither accurate nor precise as the average of the values is not close to the true value of g which is 10ms-2. 6 B Velocity is the gradient of displacement time graph. As the car goes down the slope, the velocity increases, thus the gradient of the graph should be increasing. Along the horizontal path, the velocity remains the same, thus the gradient remains constant. 8 C Acceleration is the due to the resultant force acting on the object (F=ma). Since the only force on the object is weight, the acceleration remains at g throughout the motion. 12 D By F=ma, a constant net force F, means a constant m and a. With a constant a, there must be a change in velocity. 14 B Draw free body diagrams for each individual box N and M. Apply F = ma for each box. For box M, FN on M = 3a For box N, 70 – FM on N = 7a 70 – 3a = 7a 10a = 70 a = 7 ms-2 Net force on box N = 7a = 7 x 7 = 49 N 27 A The rock possesses the most energy at the top of hill as it has GPE. At any other point, it has energy transferred to the surroundings or work done from the surroundings.

2019 End-of-Year Examination Secondary Three 2 Nanyang Girls’ High School Setters: HWQ & SL Physics II Paper 2 Section A (40 marks) 1 (a) (i) −0.03 cm (ii) diameter = 4.06 + 0.03 = 4.09 cm cross-sectional area = π(d/2)2 = π(4.09/2)2 = 13.1 cm2 (b) Rotate the vernier caliper to obtain different readings of the diameter at the same point and take the average. Measure the diameter at different points along the whole length of the rod and take the average. 2 (a) v2 − u2 = 2as v2 = 2(10.0)(4.5) + (2.0)2 v = 9.7 m s−1 (2 sf) (b) at = v − u (10) t = 9.695 – 2.0 t = 0.77 s (2 sf) (c) The time taken would be the same. The gravitational acceleration experienced by both balls is the same. 3 (a) No net or resultant force acts on the climber. No net or resultant moment acts on the climber. (b) 4 (a) The lamp rotates clockwise and topples over. As the lamp tips, its weight gives rise to a clockwise moment about the pivot causing it to topple in the clockwise direction. 25° Weight = 600 N Normal contact force = 280 N Tension

2019 End-of-Year Examination Secondary Three 3 Nanyang Girls’ High School Setters: HWQ & SL Physics II (b) Make the base of the lamp wider/ broader. Make the base of the lamp heavier. 5 (a) Let the normal contact force exerted by wall be ND. Taking moments about E, by the Principle of Moments, sum of anticlockwise moments = sum of clockwise moments 20 × 0.35 = ND × (0.10 + 0.20) ND = 23 N (2 sf) (b) Let the normal contact force exerted by wall be NE. NE = weight = 20 N (c) The magnitude of NE would be lower as NE is now equal to the difference between weight and friction. 6 (a) A point ‘S’ labelled anywhere along the bottom of the manometer. (b) Patm = hpg 0.75 x 13 600 x 10 = hwater x 1000 x 10 hwater = 10.2 m H2O Pgas = Patm + 50 cm H2O Pgas = 10.2 m + 0.50 m = 10.7 m H2O (1d.p.) or 107 000 Pa (whole number) or 110 000 (2sf) or 79 cm Hg (whole number) (c) P1V1 = P2V2 (10.7 m H2O) (0.012 cm3) = P2 (0.013 cm3) P2 = 9.9 m H2O or 99 000 Pa or 73 cm Hg (2sf) Assumption is that mass and temperature of the gas remains constant. 7 (a) The small force will exert a pressure on the hydraulic fluid through the area on the small piston. By Pascal’s Principle, this pressure is transmitted uniformly to all parts of the hydraulic fluid, thus it exerts the same pressure on the large piston. As the area of the large piston is large, due to P = F/A, the force exerted by the large piston is larger as well. This generates a large force that compresses the object. (b) By the principle of conservation of energy, the work done by the small force on the small piston must be equal to the work done by the large force on large piston. Since the force on the small piston is smaller, the distance moved must be longer for the work done to be the same. OR

2019 End-of-Year Examination Secondary Three 4 Nanyang Girls’ High School Setters: HWQ & SL Physics II The liquid is incompressible, thus the volume remains constant . Therefore, since the cross sectional area of the small piston is small, the distance moved will be larger. 8 (a) (b) The observer’s prediction is wrong. This is because image formed in plane mirrors are the same size/ smaller to the observer as the object and is upright. Section B (30 marks) 9 (a) Newton’s Second Law of Motion states that the resultant force acting upon an object is equal to the product of the mass and the acceleration of the object; the direction of the force is the same as that of the object’s acceleration. (b) 70 N weight = 20 N tension weight = 30 N tension Building Reflecting surface P I

2019 End-of-Year Examination Secondary Three 5 Nanyang Girls’ High School Setters: HWQ & SL Physics II (c) Fnet = ma 70 − 50 = (5.0) a a = 4.0 m s−2 (d) Fnet = ma T − 30 = (3.0)(4.0) T = 42 N (e) Tension would become zero. When the blocks undergo free-fall, the net force acting on each block would just be the weight of the objects. Therefore, the tension will be zero. 10 (a) (i) Using the speed of wind, The distance the air travels per second is 8.0 m. Volume of air travelling per second = Cross sectional area x distance = 𝝅 r2 x 8.0 = 𝝅 (40)2 x 8.0 = 40 200 m3 = 40 000 m3 (2sf) (ii) Density of air per second = mass / volume = 90 000 / 40200 = 2.24 kg m-3 s-1 (2 or 3 sf) (iii) rate of kinetic energy per second = ½ mv2 = ½ (90 000)(8.0)2 = 2.9 x 106 J s-1 (2sf) (iv) A large amount of air passes through the area without turning the blade OR some energy is used to do work against friction at the pivot OR the wind speed is not constant throughout the day Reject some energy is converted to sound and thermal energy of the blades, as the sound is wind and thermal energy due to friction between air and blades is too negligible as compared to the two reasons above. (b) (i) Angle of incidence = Angle of reflection = 30o (+1o) i window pane P i r

2019 End-of-Year Examination Secondary Three 6 Nanyang Girls’ High School Setters: HWQ & SL Physics II (b) (ii) The first law of reflection is that the angle of incidence is equal to the angle of reflection. The second law of reflection is that the incident ray, the reflected ray and the normal all lie on the same plane. (b) (iii) Angle of deviation = 180o – 30o – 30o = 120o 11 EITHER (a) Rate of change of velocity (b) (c) a = (v − u)/t a = (10.0 − 20.0)/2.0 = −5.0 m s−2 Deceleration = 5.0 m s−2 (d) Distance travelled by X = 0.5 (20.0 + 10.0)(2.0) + (10.0)(1.0) = 40 m Distance travelled by Y = (20.0)(1.0) + 0.5 (2