2023 XMS P2 PRELIM MS

Xinmin Sec 2023 Prelim 4E Physics 6091/02 Section A Qns Answers Remarks 1a Two vertical forces that are seemingly equal in length [B1] -- Weight -- Upthrust F/U Do not accept NCF/NRF, accept friction (with water) or (water) resistance Accept reasonable symbols 1bi The weight of the block is equal [A1] to the weight of liquid displaced by the block. Upthrust = Weight of block (from 3a) Upthrust = Weight of liquid displaced (from question) So, W block = W liquid displaced 1bii The weight of water displaced is equal [A1] to the weight of diesel displaced. W block in water = W block in diesel So, W water displaced = W diesel displaced 1c W water displaced = W diesel displaced m water x g = m diesel x g [½ for W = mg] ρ water x V water = ρ diesel x V liquid D [½ for m = ρV] ρ water x 75% x V block = ρ diesel x 86% x V block [M1] (1000) (0.75) = (ρ diesel) (0.86) ρ diesel = 872 kg/m3 [A1] From 3bii g will cancel off [1 for relating Vliquids to Vblock] Vblock will cancel off FULL MARKS IF NO WORKING BUT CORRECT ANSWER Total 6m 2ai direction of over-turning moment is clockwise [A1] 2aii clockwise moments = (U x 36) + (T x 20) = (400k x 36) + (45k x 20) = 15 300 kNm [M1] anti-clockwise moment = (W x 20) = (400k x 20) = 8000 kNm resultant moment = 15 300 – 8000 = 7300 kNm [A1] 2b Put the heavier cargo at the lower decks and lighter cargo at the upper decks. [A1] Lowers the CG Total 4m 3a P = ρgh = 1000 x 10 x 2 = 20 000 Pa [M1] F = PA = 20 000 x 0.02 [M1] = 400 N [A1] 3b Upward forces = Downward forces Normal contact forces by slabs = Force by water + Weight = 400 + 19 [M1] = 419 N [A1] upwards [A1] Total 6m 4a - The water at the bottom, nearest the heater is heated and expands. - This water becomes less dense and rises. - The comparatively cooler and denser cooler water at the top is displaced and sinks to come near the heater. - The process repeats and convection currents are set up. Any 2 points for [1m] 4b [A1] Styrofoam / cork / rubber / plastic [B1] This material contains air pockets. Air/rubber/plastic is a poor conductor of heat, so heat loss by conduction is reduced. Accept any possible answers. 4c [A1] White and smooth/shiny/glossy [B1] Light colours and smooth textures are poor emitters of heat, so heat loss by radiation is reduced. Accept any light colours besides white. Total 6m

5a 30°C --- 0.5 mV 66°C --- 0.5 / 30 x 66 [M1] = 1.1 mV [A1] 5b The higher the temperature T, (the lower the resistance of the thermistor) the smaller the potential difference across it. [A1] (emf divides amongst resistors in series proportionately) Accept opposite / vice versa. Total 3m 6a 50 complete waves are produced in 1 second. [B1] 6b 5 waves – 30 m ➔ 𝜆 = 6 m [M1] v = f 𝜆 = 50 x 6 = 300 m/s [A1] 6c The more shallow the water / the smaller the depth of water (the smaller the wavelength) the slower the wave / the lower the speed of the water waves. [A1] 6d - keep the shape of the wave and draw a stepped difference between UV and YZ (not sloping but horizontal and higher) OR - keep the shape of pool and draw the waves within YZ as becoming closer and closer [B1] Total 5m 7a critical angle ∠c = 40° [M1] n = 1 sin ∠𝑐 = 1 sin 40 = 1.56 [A1] 7b angle of refraction ∠𝑟 = 90 – 70 = 20° [M1] n = sin ∠𝑖 sin ∠𝑟 ➔ ∠𝑖 = sin-1 (n sin ∠𝑟) = 32.1° [A1] Total 4m 8ai [B1] When peeled, electrons transfer from side M to side N. [B1] Side M loses electrons and becomes positively-charged; side N gains electrons and becomes negatively-charged. Accept opposite ie from N to M 8aii [B1] Side M and side N are of opposite charges. Since unlike charges attract, the cling wrap can stick onto itself. Accept opposite ie M - ve, N +ve 8b [A1] OR [B1] The charged cling wrap can attract a neutral bowl. Accept possible alternatives 8c Make the cling wrap less taut OR Use a bigger bowl Accept possible alternatives Total 6m 9a X area of rod : X area of pipe π ( 𝐷 2)2 : π ( 2𝐷 2 )2 − π ( 𝐷 2)2 𝐷2 4 : 𝐷2 − 𝐷2 4 𝐷2 4 : 3 x 𝐷2 4 1 : 3 [A1] OR 0.333 [1 for changing diameter to radius] [1 for squaring radius to find area] FULL MARKS IF NO WORKING BUT CORRECT ANSWER

9b 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑖𝑡𝑦 × 𝑙𝑒𝑛𝑔𝑡ℎ 𝑎𝑟𝑒𝑎 rod pipe R of rod = 𝜌×𝐿 𝐴 R of pipe = 𝜌×2𝐿 3𝐴 = 2 3 × 𝜌×𝐿 𝐴 = 2 3 × R of rod Current through pipe = 3 2 x Current through rod = 3 2 x 6 [M1] = 9 A [A1] Same material = same resistivity Length x 2 → higher R, lower I by 2 Area x 3 → lower R, higher I by 3 Allow ecf FULL MARKS IF NO WORKING BUT CORRECT ANSWER Total 5m 10a [B1] The fan does not have a metal casing that can conduct electricity to its user. OR The fan has a plastic casing that can insulate electricity from its user. 10bi [B1] wire with switch to be connected to ‘live’ and wire without switch to be connected to ‘neutral’. 10bii [B1] in the live wire and between the switch and the pin-hole. 10c energy = P x t = 3 kW x 6 h = 18 kWh per day [M1] cost = 18 x 7 x $0.30 = $37.80 [A1] Total 5m Paper 2 Section B Qns Answers Remarks 11a [A1] AC generator [B1] kinetic energy into electrical energy 11b [B1] for correct circuit symbol and parallel connection [B1] for correct labelling of terminals (clue was in Fig 11a, circled in pink) Electron flow is opposite of conventional current flow. 11c maximum power Pmax = Vmax x Imax = 27.3 x 5.5 [M1] = 150 W [A1] 11di open circuit voltage Voc = 32 V When current is zero 11dii short circuit current Isc = 5.5 A When pd is zero 11e irradiance of 1000 W/m2 means there is [B1] 1000 J of (light) energy [B1] per second (unit time), per square metre (unit area) Accept explanation by both quantities or by units Total 10m

12a 70 km/h = 70 000 m / 3600 s [M1] = 19.4 m/s [A1] 12b Distance travelled = Area under v-t graph [M1] = (20x40) + ½(20+30)20 + (30x60) + ½(30+10)20 + (10x40) + ½(10+20)20 + (20x20) = 800 + 500 + 1800 + 400 + 400 + 300 + 400 = 4600 m [A1] 12c Average speed = total d / total t = 4600 / 220 = 20.9 m/s [A1] [B1] The car was speeding as its average speed of 20.9 m/s is higher than speed limit of 19.4 m/s. Allow ecf Allow ecf 12di WD = F x d = P x t ➔ F = Pt/d [M1] Driving force = 40 000 x 2 /15 = 5333 ≈ 5330 N [A1] 12dii Initial energy = final energy KEi + GPEi = Kef [M1] (½ x 1200 x 152) + (1200 x 10 x 3) = (½ x 1200 x v2) v = √285 =16.88 ≈ 16.9 m/s [A1] Total 10m EITHER 13ai Power in P = I2R = (2.52)(28.0) = 175 W [A1] Power in Q = I2R = (2.02)(28.0) = 112 W [A1] 13aii Difference between P & Q: The rate of temperature rise in P is greater than the rate of temperature rise in Q. Explanation: The rate of input of thermal energy in P is greater than the rate of input of thermal energy in Q. [1] [1] 13b Relationship between θ and t: (In both P & Q) As t increases, θ increases at a decreasing rate. Explanation: - As t increases, temperature of water increases. The temperature difference between the water and the surroundings increases. - The rate of heat loss to the surroundings increases, so temperature rise decreases. [1] [1] [1] 13ci Q = mcθ → Q/t = mcθ/t → P = mcθ/t [B1] 13cii θ/t is gradient of graph Power in P = mc(θ/t) = mc(gradient) 175 = (m)(4.2)(12 K min-1) = (m)(4.2)(0.2 K s-1) [M1] m = 208.3… = 208 g [A1] Allow ecf Total 10m OR 13ai X 13aii Y and Z 13b When V1 is 8 V, V2 is 16 V. 𝑁𝑠 𝑁𝑝 = 𝑉𝑠 𝑉𝑝 𝑁𝑠 = 𝑉𝑠 𝑉𝑝 × 𝑁𝑝 = 16 8 × 250 = 500 [M1A1] 13c [B1] Some energy is lost as thermal energy due to the resistance of the coils. OR [B1] Some energy is lost due to leakage of magnetic field lines between the primary and secondary coils. OR Any two out of fo