SJI_2020 Year 3 EOY Physics 6091 ANSWERS

SJI 2020 Y3OP PHYSICS EOY EXAM MARKING SCHEME Paper 1 1 B 11 C 21 B 2 D 12 C 22 B 3 B 13 B 23 B 4 A 14 D 24 D 5 A 15 D 25 A 6 B 16 A 26 B 7 B 17 B 27 C 8 A 18 A 28 B 9 B 19 D 29 A 10 B 20 D 30 B Paper 2 Section A 1 a. t100 = 3.00 mm + 0.37 mm = 3.37 mm t = 3.37 / 100 x 10-3 m = 3.37 x 10-5 m bi. thickness= 0.22 mm = 0.00022 m volume = 0.00022 m x 0.594 m x 0.841 m = 0.0001099019 m3 = 0.00011 or 0.000110 m3 bii. Mass of A1 cardboard = 400g x (0.594 m x 0.841 m) = 200 g ρ = mass / volume = 0.200 g / 0.00011 m3 = 1820 kg/m3 or 1800 kg/m3 2 a. a = v-u/t = (30-10)/(3.0-1.0) = 10 m/s2 b. From 3.0s to 4.0s : Velocity is decreasing at a constant rate/ uniform/constant deceleration and the jumper comes to rest momentarily at 4.0 s. From 4.0s to 5.0s: Velocity is increasing at a constant rate/uniform/constant acceleration as the jumper moves up/rebound upward. ci. Displacement = (½ x 30 x 4) – (1/2 x 2 x 8) = 52 m cii. Average speed = (½ x 30 x 4 +1/2 x 2 x 8) / 6.0 s =11.3 or 11 m/s

ecf applies ciii. The distance travelled by the jumper is greater than his displacement as it does not take into account the direction of motion of the jumper (as such average speed is greater than the average velocity) 3 a. gas P = pgh = 13 600 x 10 x 0.40 = 54 400 or 54 000 Pa b. Cabin air pressure = 40 + 20 = 60 cm Hg c. h1 drops while h2 increases/the difference between h1 and h2 decreases. This is because the difference between the gas pressure and the cabin air pressure decreases. Or As the cabin air pressure increases and the gas pressure remains constant. The pressure on the left limb becomes higher than the right limb, creating a net force which pushes the mercury (down on the left limb and up on the right limb). di. At high altitude the height of the layer of atmosphere above the aircraft is smaller / at high altitude the density of air is lower/the gravitaional field strength is lower. accept: KMT answer less particles per unit volume/lower temperature→ less frequency/force of collisions → less force/unit area→ less pressure dii. P = F x A = 80 000 Pa x 0.0875 m2 = 7000 N [-1M] for wrong conversion of area from cm2 to m2 4 a. Loss in GPE is equal to the gain in KE of toy car and gain in thermal energy of the toy car and surroundings due to work done against friction. The sum of the mechanical energy and thermal energy is always constant. bi. 1.0 kg x 10 N/kg x 0.30m = 3.0 J bii. Loss in GPE = gain in KE + work done against friction 3.0 J = 0.50(1.0kg)(2.5m/s)2 – 0.50(1.0kg)(u2) + 0.11N(0.90m)

u = 0.67 m/s c The height and frictional force remain unchanged, hence the change in KE is unchanged. Doubling the initial speed would not double the final speed. 5 a. Air