2024 MSH SC(PHY) PRELIM ANS

MARIS STELLA HIGH SCHOOL SECONDARY 4 SCIENCE(PHYSICS) 2024 PRELIMINARY EXAMINATION Deduction of 1 m per occurrence (cap of 1 m per type of error for the entire paper): • Incorrect or unclear presentation (P) or Incorrect or missing formula (F) • Incorrect or missing units (U) • Incorrect no. of significant figures (SF) Paper 1 [20 marks] 1 2 3 4 5 6 7 8 9 10 D A D A A C A D A D 11 12 13 14 15 16 17 18 19 20 C C B B B B A C B C Paper 2 Section A [55 marks] Answer all the questions in this section. Qn Mark Scheme Marks 1 (a) Best fit curve from t = 0 s to t = 26 s [B1] Straight line through all 3 points from t = 26 s to t = 30 s [B1] (b) (i) 𝐸𝑘 = 1 2 𝑚𝑣2 = 1 2 (46 𝑘𝑔)(15.2 𝑚 𝑠⁄ )2 = 5313.92 𝐽 = 5300 𝐽 (𝑡𝑜 2 𝑠. 𝑓. ) C1 A1 (ii) 𝑎 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑣 − 𝑡 𝑔𝑟𝑎𝑝ℎ = 0 − 15.2 30 − 26 = −3.8 𝑚 𝑠2⁄ deceleration = 3.8 m/s2 C1 A1 (iii) 𝐹 = 𝑚𝑎 = (46 + 9 𝑘𝑔)(3.8 𝑚 𝑠2⁄ ) = 209 𝑁 = 210 𝑁 (𝑡𝑜 2 𝑠. 𝑓. ) C1 A1

2 (iii) Energy is transferred from the kinetic store and gravitational potential store of the child and sledge to internal energy of surroundings through a mechanical pathway. B1 B1 2 (a) (i) 𝜌 = 𝑚 𝑉 1000 𝑘𝑔 𝑚3⁄ = 𝑚 2.5 × 10−4 × 0.12 m3 m = 0.030 kg (F) C1 A1 (ii) 𝑃 = 𝐹 𝐴 = (0.030 𝑘𝑔 × 10 𝑁 𝑘𝑔⁄ ) 2.5 × 10−4 𝑚2 = 1200 𝑁 𝑚2⁄ (F) C1 A1 (iii) 𝑃 = 𝐹 𝐴 1200 𝑁 𝑚2⁄ = (𝜌 × 2.5 × 10−4 × 0.15 m3 × 10 𝑁 𝑘𝑔⁄ ) 2.5 × 10−4 𝑚2 𝜌 = 800 𝑘𝑔 𝑚3⁄ C1 A1 (c) Turn the clamp so that the U-tube is directly above the base of the retort stand OR Lower the clamp so that the centre of gravity of the set up is lowered. B1 3 (a) (i) newton-meter OR spring balance (ii) measuring tape B1 B1 (b) (i) The principle of moments states that when a body is in equilibrium, the sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot B1 B1 (ii) In equilibrium, taking moments about the pivot, Sum of clockwise moments = sum of anti-clockwise moments 𝐹 × 5.000 𝑚 = 52.0 𝑁 × 0.250 𝑚 𝐹 = 2.6 𝑁 (P) C1 A1 4 (a) Heat is transferred via conduction Copper is a good conductor of heat as it has free electrons. B1 B1 (b) The water around the copper pipe is cooled, contracts and becomes denser, allowing it to sink to the bottom. Warmer water rises to take its place forming convection currents. Since water is a poor conductor of heat, t his cools the water below efficiently allowing less dense room temperature water to remain above. B1 B1 (c) As temperature of the liquid decreases, the molecules lose average kinetic energy, they slide over one another at lower speeds. B1 B1 5 (a) (i) 340 m/s A1 (ii) 𝑣 = 𝑓𝜆 340 𝑚 𝑠⁄ = 485 𝐻𝑧 × 𝜆 𝜆 = 0.70 𝑚 (F) E1 A1 (iii) Sound waves of different frequencies travel at the same speed in the same medium. B1

3 (b) As speed of sound in metal is greater than in air, the sound will take a shorter time to travel the same distance and hence timer will have a smaller reading. B1 6 (a) The fur lost negative charges to become positively charged. The plastic gained negative charges to become negatively charged. B1 B1 (b) (i) 𝑊 = 𝑚𝑔 = 0.015 𝑘𝑔 × 10 𝑁 𝑘𝑔⁄ = 0.15 𝑁 (F) C1 A1 (ii) force arrows act from centre, in opposite directions but same magnitude and correctly labelled [B1] 7 (a) Q, +1.6 × 10-19 C or +1 charge B1 (b) Q and S B1 (c) (i) R and S B1 (ii) 99 years = 3 × 𝑡1 2⁄ 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = 1 2 × 1 2 × 1 2 = 1 8 𝑂𝑅 0.125 C1 A1 (iii) β-particles can pass through the paper depending on its thickness, while α-particles would be blocked and γ-particles can pass through the paper freely. B1 8 (a) (i) Current is the rate of flow of electric charge. B1 (ii) 𝐼 = 𝑉 𝑅 𝐼 = 12 𝑉 20 + 28 Ω = 0.25 𝐴 (F) C1 A1 (iii) The magnetic field due to the current interacts with the magnetic field of the permanent magnets. This causes a stronger field to be set up above the wire and weaker field set up below the wire. Hence YZ experiences a downward force from stronger to weaker field. B1 B1

4 (b) (i) ( 1 30 + 1 20) −1 = 12 Ω 12 Ω + 28 Ω = 40 Ω C1 A1 (ii) When S2 is closed, current through the wire increases, causing the force on YZ to increase in magnitude OR become stronger. B1 B1 Section B Answer one question from this section. 9 (a) 𝑣 = 𝑓𝜆 3.0 × 108 𝑚 𝑠⁄ = 𝑓 × 589 × 10−9 𝑚 𝑓 = 5.093 × 1014 𝐻𝑧 = 5.1 × 1014 𝐻𝑧 (𝑡𝑜 2 𝑠. 𝑓. ) (F) C1 A1 (b) (i) 𝑛 = sin 𝑖 sin 𝑟 1.5 = sin 60° sin 𝑟 𝑟 = 35° (𝑡𝑜 2 𝑠. 𝑓. ) (F) C1 A1 (ii) Drawn using ruler and protractor where r matches answer in (b) [B1] (iii) 𝑛 = 𝑐 𝑣 𝑣 = 3.0 × 108 𝑚 𝑠⁄ 1.5 = 2.0 × 108 𝑚 𝑠⁄ (F) C1 A1

5 (c) Correct ray diagram (any 2 correct rays with direction) [B1] Accept f = 16 – 19 cm [B1] (ii) The focal length of the lens decreases. B1 10 (a) (i) 𝑃 = 𝑉𝐼 3000 𝑊 = 240 𝑉 × 𝐼 𝐼 = 12.5 𝐴 (F) C1 A1 (ii) 13 A E1 (iii) 𝑐𝑜𝑠𝑡 = 𝑃𝑡 × 𝑟𝑎𝑡𝑒 = 3 𝑘𝑊 × 10 60 ℎ × 30 × $0.2989 = $4.48 (𝑡𝑜 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑐𝑒𝑛𝑡) (F) C1 A1 (b) (i) wire X: neutral wire Y: live wire Z: earth B1 B1 B1 (ii) The earth wire is connected to the live terminal. This causes the metal casing to be live and anybody who touches it can get an electric shock. B1 B1