Beatty 2026 Physics P1 & P2 MS

Paper 1 BEATTY SECONDARY SCHOOL PRELIMINARY EXAMINATION 2026 SECONDARY FOUR EXPRESS / G3 Pure Physics 6091 Mark Scheme Papers 1 & 2 1 A 11 B 21 C 31 D 2 A 12 B 22 C 32 A 3 B 13 B 23 D 33 C 4 B 14 D 24 D 34 D 5 A 15 B 25 D 35 C 6 C 16 C 26 C 36 A 7 D 17 A 27 C 37 B 8 C 18 C 28 C 38 A 9 A 19 A 29 A 39 B 10 D 20 D 30 C 40 C

Paper 2 Section A Qn Answer Mark Remark 1a acceleration = (5-0)/(2×60) = 0.042 m/s2 (2 s.f.) [1] [1] 1b Monorail undergoes constant deceleration until it momentarily stops at 7.5 minutes. It continues to accelerate in the opposite direction with the same constant magnitude. [1] [1] 1c Distance from starting point = Area under graph = ½ (5)(3+5)(60) = 1200 m [1] [1] 1d [1] – correct shape in first 7 seconds, all times labelled [1] – correct shape in last 3 seconds, graph does not return to zero 2a [1] [1] 2b Weight of student and the gravitational pull of the Earth by the student OR Tension in rope due to the student’s pull and the force exerted on the student by the rope. [1] displacement / m time / min 0 2 7 8 7.5 10 W 620 N T F 24° F = 276 N T = 679 N Scale = 1cm: 100 N

2c As the angle decreases, the tension T is expected to increase while reaction force F will decrease. [1] 3ai Weight of lorry tyre = (75)(10) = 750 N Pressure in oil = 750 / 0.960 = 780 Pa (2 s.f.) [1] [1] 3aii Force on piston A = 782 (0.080) = 62.5 N Using POM, ACWM = ACM F(60) = 62.5 (10) F = 10.4 N [1] [1] ecf 3b Oil is in the liquid state, in which particles are closely packed with little or no space in between. This makes oil incompressible and pressure throughout the oil remains the same. Oil particles also slide past one another, thus enabling oil to flow and take the shape of its container. [1] [1] [1] 3c When oil can flow freely through the release valve, pressure in the oil cannot be effectively transmitted since oil at piston B is exposed to the atmosphere via the oil reservoir. The lorry tyre will not be able to be lifted when the handle is pressed, nor when the handle is pulled up. [1] [1] 4ai 1300 kg/m3 [1] 4aii Pressure of trapped air = atmospheric pressure + pressure due to column of liquid = 1.01 × 105 + hρg = 1.01 × 105 + (0.480)(1300)(10) = 107 240 = 107 kPa (3 s.f.) [1] [1] 4bi Glass particles nearest the warm liquid gain energy and vibrate more vigorously. They then collide with neigbouring glass particles, causing them to vibrate more vigorously too. As kinetic energy is transferred across the glass, the liquid decreases in average kinetic energy and cools until thermal equilibrium is reached. [1] [1] 4bii x will decrease when glass tube is in cold room. Space between air particles will significantly decrease, causing a reduction in volume occupied. [1] [1]

4biii Density of liquid increases as the liquid contracts. Since hydrostatic pressure difference (hρg) remains the same, height h is expected to decrease as density ρ increases. [1] [1] 5ai Energy cannot be created nor destroyed. It can only be transferred from one store to another. The total amount of energy in an isolated system remains constant. [1] 5aii At the top of the cliff, the stationary pebble has only energy in the gravitational potential store. This is the total amount of energy in the system. As the pebble drops, since energy cannot be created or destroyed, part of the energy in the gravitational potential store is transferred to the kinetic store, making the pebble drop faster. Just before hitting the bottom, almost all of the energy in the gravitational potential store has been transferred to kinetic store. Pebble is at its maximum speed. [1] [1] [1] 5b Using PCOE, Total energy at top = total energy at bottom GPE = KE m(10)(380) = ½ (m)(v2) v = 87.2 m/s [1] [1] 5c 1. Height of cliff is a quarter lower than the earlier cliff. Ignoring air resistance, height of this cliff can be calculated using PCOE. GPE = KE m(10)(h) = ½ (m)(87.2 / 2)2 h = 95 m (i.e. ¼ of 380 m) OR 2. Wind or updraft or air resistance could have reduced the maximum speed of the pebble. No change to the cliff height. Cliff height = area under graph 380 = ½ (t)(87.2) Original time taken t = 8.72 s [1] [1] [1] OR [1] [1] time/s velocity /ms-1 0 87.2 8.72 13.08

380 = ½ (8.72)(u) + (u)(8.72 / 2) u = 43.6 m/s (i.e. ½ v) [1] 6a density = mass / volume 1.04 = mass / 150 mass = 156 g [1] 6b Qsupplied = mscsΔθs + mlv + mwcwΔθw = m(2200)(120-100) + m(2260000) + m(4200)(100-60) = 2,432,400m Qgain = mmcmΔθm = (0.156)(3900)(60-4) =34070.4 J Qsupplied = Qgain 2,432,400m = 34070.4 m = 0.0140 kg [1] [1] [1] 6c Heat from the steam is transferred via infrared radiation to the cold milk. Heat is also transferred via conduction from the steam bubbles to the cold milk. [1] [1] 7a n = sin i / sin r = (sin 27) / (sin 17) = 1.55 [1] [1] 7bi Need to show that: 1. R bends in the prism 2. Focal length is now further [1] 7bii Any two of the following: 1. Radio waves 2. Microwaves 3. Infrared waves [1] 7biii Speed of light in a given medium is inversely proportional to its refractive index (n = c / v). Since red light has lower refractive index than blue light, its speed is thus greater than that of blue light. [1] [1] R

8ai resistance = 200 Ω [1] 8aii temperature = 50 °C [1] 8bi effective circuit resistance = (1/100 + 1/1000)-1 + 200 = 290.9 Ω Using Ohm’s Law, I = 6.0/290.9 = 0.021 A (2 s.f.) [1] [1] Ecf from 8ai 8bii Effective resistance of 100 Ω and 1000 Ω resistors = (1/100 + 1/1000)-1 = 90.9 Ω Using potential divider concept, V = 90.9 /(90.9 +70) × 6.0 = 3.39 V Hence, P = V2/R = (3.39)2/1000 = 0.011 W (2 s.f.) [1] [1] Ecf from 8aii 8c Increase in temperature will lead to a decrease in resistance in the thermistor. Since the thermistor and the 100 Ω resistor are in series (potential divider circuit), the potential difference across the lamp will be higher. With a higher potential difference across the lamp, the power output is greater and the lamp is brighter. [1] [1] [1] 9a The emission of radiation by a radiactive nuclide is unpredictable and it is not possible to determine when it will happen. [1] 9bi Alpha particle [1] 9bii Po84 210 → Pb82 206 + α2 4 [1] 9biii [1] 9c Initial count rate without background radiation = 20736 Count rate after 140 days with out background radiation = 10368 Ratio = 20736 / 10368 = 2 Hence, half-life of polonium-210 = 140 days [1] [1]

9di Destroying cancerous cells in radiotherapy during cancer treatment. OR any other medical treatment [1] 9dii Half-life: Iridium-192 should have a shorter half-life so that it makes it safer to handle. Penetrating ability: Due to presence of gamma emission that is highly penetrative, this isotope can be used to penetrate deeper and target a specific spot in the human body. Ionising ability: Iridium-192’s gamma rays also have low ionising ability, making it less likely to harm healthy cells in the vicinity of the cancerous cells. [1] [