NYGH-2023-S4-EOY-Physics-Answers

Answers for 2023 Sec 4 IP Physics EOY Exams Paper 1 1…….5 6…...10 11…..15 16….20 21….25 26….30 AABCB DCDBC ADDDC BABBA BDABC CCADA Paper 2 Section A 1(a)(i) Velocity is a vector quantity whereas speed is a scalar quantity Or Velocity has magnitude and direction whereas speed has magnitude only. (a)(ii) Change in velocity = 20− (−8) = 28 m s−1 (a)(iii) Change in speed = 20 − 8 = 12 m s−1 (b)(i) At max height, v = 0 m s−1 Displacement = area under the v-t graph (up to t = 0.80 s) = ½ (0.80)( −8.0) = - 3.2 m (b)(ii) Displacement = Area under the v-t graph = ½ (2.0)(20) – 3.2 = 16.8 m (b)(iii)  correct shape  correct labels (allow e.c.f. for displacement values) 2(a) (i) A measure of the amount of substance in an object (ii) The gravitational force on the object (b) Taking pivot at B, applying Principle of moments, Total clockwise moments = total anticlockwise moments 50 (0.30) + 25 (0.30 + 0.60) = FA (0.80 – 0.30) Evidence of M = Fd applied correctly for either side of the equation FA = 75 N (c) Total upward forces = total downward forces FB = 75 N + 50 N + 25 N = 150 N 3(a) density = mass / volume = (66.45 – 52.00) / 17.0 = 0.85 g cm–3 (b) hw ρw g = hρXg  (12.0 – 5.0) (1.0) = h (0.85) h = 8.2 cm 4(a) (i) radiation (ii) conduction (b) the black surface is a good absorber of infrared radiation from the sun. (c) Copper is a better thermal conductor to transfer thermal energy to the water displacement / m time / s 16.8 0.80 1.60 2.80 -3.2

2 5 Let t be final temperature of coffee loss of heat by steam + loss of heat by water = heat gained by coffee to become water M L + M c Δθ = m c Δθ (0.025 × 2.26 × 106) + 0.025 × 4200 × (100 - t) = 0.15 × 5800 × (t - 20) 5.65 × 104 + 10500 - 105 t = 870 t - 17400 t = 86.56  87 C 6(a) Draw an additional light ray through the optical centre to intersect original light ray to locate the image position. Image distance = (12÷2) × 10 cm = 60 cm. (b) Draw an additional light ray horizontally, then to meet the image. Focal length f = (4÷2) × 10 cm = 20 cm. OR calculation using 1/f = 1/u + 1/v 7(a) A current flows through him from the car to earth to discharge the net charge on the car body. OR his body is a conductor, so net charge on the car can be discharged by earthing. (b) Use conducting / conductive material / material which is a good electrical conductor / special conducting rubber with high carbon content to make the car tyres so any excess charge on the car can be removed by earthing. 8(a) Area =  r2 =  × (d/2)2) =  × (0.13/1000)2 = 5.309 ×10-8 m2 R =  L / Area = 1.7 × 10-8 m × 1.0 m / 5.309 ×10-8 m2 = 0.32  (b) resistance of 1 strand of 5 m = 0.32  × 5.0 = 1.6  2