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 particles move in continuous and random direction. The air particles constantly collide with the wall of the container and since there is area of contact, pressure is exerted on the wall. b. Temperature drops and the average KE decreases. Frequency and force of collision with the wall decreases. Internal P decreases and is lesser than the external water+atm pressure. As such net force acting inward. or The vaporised water particles condenses and the number of air particles per unit volume in the can will decrease. Frequency of collision with the wall decreases. Internal P decreases and is lesser than the external water+atm pressure. As such net force acting inward. 6 ai. Vibration and collision of particles Free electron diffusion aii. Cooler denser water sinks, warmer water below rises, set up convection currents result in mixing. 7 a. It is a physical property that changes linearly and continuously with temperature over a range of temperature. b. 𝜃 = 30𝑘𝑃𝑎−10𝑘𝑃𝑎 90𝑘𝑃𝑎−10𝑘𝑃𝑎 × 100℃ = 25℃ 8 Light travels in straight line from lamp to A and P. Light reflecting off mirror is also incident on P but not A as the length of the mirror is not sufficient for light to reflect to A.

Section B 9 ai. Fr = ma 800 N – f = (1200kg) x (0.50 m/s2) f = 200 N aii. The velocity increases and the air resistance increases. The resultant force decreases, resulting in the acceleration decrease. bi. 1500 N (or tension) as the tension on the boat is the equal and opposite to the tension on the man. bii. Correct shape correct arrows 2200 ± 250 N biii. Fr = ma 1500 N = (100 kg)(a) a = 15 m/s2 Directly is opposite to the tension of the rope on the man 10 ai. The centre of gravity of a body is a point through which its whole weight appears to act for any orientation of the body. aii The cardboard will rest when its centre of gravity is vertically below the pivot P. At this position, (there is no perpendicular distance between its weight and the pivot point) the cardboard is not having any moment about the pivot. bi

bii At equilibrium, the sum of clockwise moments about A = sum of anti- clockwise moments about A (W x 2.0 cm) + (1.3 N x 4.0 cm) = 1.0 N x 6.0 cm 2.0W + 5.2 Ncm = 6.0 Ncm 2.0W = 0.80 Ncm W = 0.40 N biii Mass of rod = 0.40 N / 10 N/kg = 0.040 kg biv. Sum of forces acting up = Sum of forces acting down TA + TB = 0.40 N + 1.3 N (accept ecf of W from 2bii.) TA = 0.40 N + 1.3 N – 1.0 N TA = 0.70 N bv. M has to move to the left. So that the sum of clockwise moments of M about A is reduced, and hence tension B can also be reduced to equalize tension A. 11 ai. Ratio of speed of light in vacuum to speed of light in glass is 1.6 aii. 32.8o aiii. 1 38.7o aiii. 2 Angle of incidence >critical angle hence TIR occurs. aiii. 3 TIR and emerge at 60o to normal aiv. r increases. Light slows down less when travelling from water to glass as the ratio of the refractive indices decreases. bi Due to the curved surface of the lens, light rays incident near the edges have a larger angle of incidence hence the light rays would refract by a larger angle. bii 1 Image height twice object height, arrows bii 2 v = 36 cm