NYGH-2015-S3EOY-Physics P1 and P2 Ans

2015 Sec 3 Physics EOY Answers (Students’ Copy) Paper 1 (30): 1 ..…. 5 6 …… 10 11 ..…. 15 16 ..…. 20 21 ..…. 25 26 ..…. 30 CBCAD CACBC DDCDA CCCBC CDBCB BAADA Solutions to selected questions: 7: The object attains terminal velocity (constant velocity) means there is air resistance which would reduce the acceleration from 10 m s-2 till 0 m s-2 (when net force becomes zero). 15: The fan blows air towards the right, the air exerts an equal and opposite force on the fan towards the left. This air also exerts a force on the board towards the right. There is no net force on the fan & board (both mounted on the cart). 17: Apply Newton’s 3rd law of motion. (3) Your weight is the gravitational force acting on your body by Earth, while normal contact force is the force acting on your body by the ground. Both forces act on the same body, so they are not a pair of action-reaction forces. 21: 1st cube: P1 = W/A = mg/A = Vρg/A = x3ρg/x2 = xρg ; P2 =(2x)3ρg/(2x)2 = 2xρg = 2 P1 24: efficiency = useful work / total energy = useful work / (P x t), t = 3600 s. Paper 2 Section A (40 marks) 1(a) 1.10 kg [1] 1(b) Mrs Tan’s method is more accurate. [1] Each time a measurement is taken with the weighing scale, an error of ± 50 g could be made. [1] EITHER weighing n number of fruits separately could result in an error of ± n × 50 g, hence, it is better to weigh all the fruits at once. [1] OR The mass of more fruits weighed together is much greater than a single fruit. The error of ± 50 g hence, becomes less significant. [1] 2(a) a = (v – u)/t = (46.4 – 0) / 10.0 [1] = 4.639 ≈ 4.64 m s−2 [1] (b) t = distance / v = 500 / 46.39 [1] = 10.78 ≈ 10.8 s [1] (c) • Shape of graph [1] • Labelling of axes, speed and times for each stage [1] 3(a) P = Fnet = ma = (4.0 + 3.0 + 2.0) × 2.1 = 19 N [1] 3(b) block A: P – FA = ma  18.9 - FA = 4.0 × 2.1 [1] Force on A by B, FA = 10.5 N [1] OR FA =force on B by A = Fnet = ma = (3.0 + 2.0) × (2.1) 3(c) Net force on C, Fnet = ma = 2.0 × 2.1 = 4.2 N [1] speed / m s−1 time / s 46.4 0 10 20.8 40.8

4(a) “weight” (or gravitational force) & “normal contact force” [1] • Correct direction of arrows 4(b) OR • Correct shape & orientation of triangle [1] • Forces labelled (or with suitable symbols) and at least 2 angles [1] • Correct direction of arrows [1] Note: Draw forces in same direction as original forces in the diagram! 4(c) Apply the force F parallel to the plane (or at any angle closer to the surface of the plane) at the same point on the box. [1] 5(a) For a system/body/object in equilibrium, the clockwise moments about any point must equal the anti-clockwise moments about the same point. [1] 5(b) Applying principle of moments, 3.0 × 180 = y × 70 [1] y = 7.7 cm [1