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] 5(c) Total mass of R and S = 180 + 70 = 250 g [1] R and S must be of the same mass. Therefore, mass of R = mass of S = 250 ÷ 2 = 125 g [1] 5(d) Rod 1: Rod 1 will turn in the anti-clockwise direction. [1] Rod 3: Rod 3 will turn in the clockwise direction. [1] 6(a) Pressure due to water = 40.0 × 1000 × 10 = 400000 Pa [1] Pressure in bubble = pressure due to atmosphere + pressure due to water = 400 000 + 101 000 [1] = 501 000 Pa or 501 kPa [1] 6(b) Volume of the bubble will increase. [1] As the bubble rises, the pressure in the bubble decreases, according to Boyle’s Law, the volume will then increase accordingly. [1] 7(a) Speed of car = 70 km h -1 = 70000/3600 = 19.4 ms-1 [1] Kinetic energy of the car = ½ (1800)(19.4)2= 338 724 J = 339 kJ [1] 7(b) Distance travelled by car in 1 min = 70 / 60 × 1000 = 1167 m [1] Or 19.4 × 60 = 1164 m Work done against friction = F × d = 400 × 1167 = 466 667 J = 467 kJ (3 s.f) [1] 7(c) Power required = Work done / time = 466 667 J/ 60 s = 7778 W = 7.78 kW [1] Weight or W normal contact force or N F 30° W N F 30°

8(a) Distance of object in front of mirror = distance of image behind mirror. Correct image, laterally inverted. All construction lines must be drawn. [2] 8(b) Light rays with arrows in correct direction, going into observer’s eye. 2 sets of rays drawn from Q to two corners of eye to image of Q. Angle of incidence = Angle of reflection [2] Paper 2 Section B (30 marks) 9(a)(i) freefall: v2 = u2 + 2as  v = √(02 + 2(10)(3.0) [1] = 7.746 ≈ 7.7 m s-1 (2 s.f.) [1] OR Loss in g.p.e. = Gain in k.e. mgh = ½ mv2  v = √(2gh) = 7.7 m s-1 9(a)(ii) on impact: v2 = u2 + 2as  a = (v2 – u2)/ 2s = (02 – 7.7462) / 2(0.65) [1] = - 46.15 ≈ - 46 m s-2 (2 s.f.) [1] deceleration = 46 m s-2 9(a)(iii) F: force on torso by his legs W – F = ma F = W - ma = m (g - a) = 45 (10 – (-46.15)) [1] = 2529 ≈ 2500 N [1] 9 (b) When the parachute opens, a large air resistance (or drag force) acts upwards on the parachute & trainee. [1] This air resistance is greater than the total weight of parachute and trainee. [1] / the net force is upwards causing him to decelerate. 9(c) As he slows down (or decelerates), the air resistance decreases. [1] When air resistance is equal to the total weight (of parachute and trainee) [1] / net force is zero / acceleration is zero, hence he reaches a constant speed. 10(a) Construction lines on a fully labelled proper diagram. Connection between diagram and subsequent working must be explicit. Diagram to show reasonable proportions. [1] Explicit and clear working with statements leading to answer of 2.4 m. Answers with insufficient working will not be given the mark. [1] W F a

10(b) There are many approaches to solving this question. Explicit and clear working with statements leading to answer of 3.26 m or 3.3 m must be shown. [2] length = 3.26 m or 3.3 m. Answers with insufficient working will not be given the mark. [1] 10(ci) clearly indicate angle i and and angle r on Fig 10.2 [1] 10(cii) Some form of Snell’s Law clearly indicated: 1.00 sin 50° = 1.5 sin r [1] Angle of refraction = 30.7 ° or 31 ° [1] 10(ciii) Glass of a higher refractive index reduces the angle of refraction, leading to a smaller angle of incidence and reflection at the reflective coating. [1] The ray from the main image will emerge closer to the ray from the ghost image, making the ghost image less distinct. [1] 11 EITHER 11(a)(i) s = 0.60 + 0.40 = 1.00 m [1] 11(a)(ii) u = 20 m s-1, v = 0 m s-1, s = 1.00 m v2 = u2 + 2as  a = (0 – 202) / 2(1.00) = - 200 m s-2 [1] average deceleration = 200 m s-2 [1] 11(b) u = 20 m s-1, v = 0 m s-1, s = 1.00 m v = u + at  t = (0 – 20) / (-200) [1] = 0.10 s [1] 11(c) F = ma = 65 x 200 = 13 000 N or 1.3 x 104 N [1] 11(d) K.E. = ½ mv 2 = ½ (65) (20)2 [1] = 13 000 J or 1.3 x 104 J [1] 11(e) K.E. of the car decreases to zero. [1] K.E. is converted to thermal (and sound) energy in the brakes which is dissipated to surroundings. (Other irrelevant energies mentioned will void the answer) [1] 11 OR 11(a) P = F / A = 25 200 / 4 x 120 x (0.01)2 [1] = 525 000 ≈ 530 000 Pa or 5.3 x 105 Pa [1] 11(b) Ballerina: P = F / A = mg/A = (56 x 10)/ (0.01) 2 = 5.6 x 106 Pa [1] Ballerina exerts a higher pressure on the ground than the elephant. [1] 11(c)(i) weight and normal contact force indicated with the length of the weight vector at least equal to the normal contact force vector. [1] 11(c)(ii) horizontal component = 1200 cos 25° [1] = 1088