XMS 2026 Physics P3 MS

Marking Scheme Physics (6091) Preliminary Examination 2026 Paper 3 Qn Marking Scheme Remarks Marker’s Report 1ai [1] l1 = answer to 3 dp, in the range of (0.090 – 0.100) h1 = answer to 3 dp, range of (0.027 – 0.033 m) Stretched and 3 cm above l1 = 0.095 m (±10%) h1 = 0.030 m 1aii [1] l2 = answer to 3 dp in the range of (0.010 – 0.240) [1] h2 = answer to 3 dp, h2 > h1 1b k = 25 N/m W = 2.0 N 1bi Eep = 0.084 J (2 sf) [1] Correct calculation [1] Correct sf with correct unit Loss in elastic store Eep = 0.5 k (l1 - l2)2 = 0.5(25)(0.095 – 0.013)2 = 0.08405 1bii Egp = 0.17 J (2 sf) [1] Correct calculation with correct sf Gain in gravitational store Egp = W (h2 - h1) = 2.0 (0.113 - 0.030) = 0.166 1c 𝐸𝑒𝑝 𝐸𝑔𝑝 = 0.51 (2 sf) [1] Correct calculation with correct sf and no units Ratio 𝐸𝑒𝑝 𝐸𝑔𝑝 = 0.08405 / 0.166 = 0.506 325 Mark scheme will use these abbreviations / + R A Ig ref ecf AW AVP ORA OWTTE underline ( ) alternatives statements on both sides of the + are needed for that mark reject accept (for answers correctly cued by the question) ignore as irrelevant with reference to error carried forward alternative wording (where responses vary more than usual) alternative valid point or reverse argument or words to that effect actual word given must be used by candidate (grammatical variants excepted) the word / phrase in brackets is not required but sets the context

1d [1] SOE 1 (about spring): • As the spring stretches, it is not possible to find the same reference points to measure length of the spring, hence there will be inaccuracies in the readings of l1 and l2. • As the spring unstretches, it may not return to its original unstretched length, hence there will be inaccurac y in the readings of l2. • The spring may not be exactly vertical, hence there will be inaccuracies in the readings of l1 and l2. [1] SOE 2 (about ruler): • The half metre ruler cannot be held steadily by hand close to the spring and the load, hence there will be inaccuracies in the readings of l1, l2, h1 and h2. • The half metre rule is not held exactly vertical, hence there will be inaccuracies in the readings of l1, l2, h1 and h2. [1] Improvement: • Use a spring with a reference wire/pin on both ends. • Clamp a metre rule vertically close to the spring. • Use a set square against the base of the retort stand and the spring/rule to ensure that spring/rule is exactly vertical. Accept: Uncertainty as to where e is zero. Affecting l2 or h2. Uncertainty in reading l 2 or h 2 when the mass is held by hand as it may unconsciously be moved. [Total: 10] 2a [1] Alignment for z-axis: Align the retort stand, block with lenses and screen along the side of the metre rule and ensure that they are perpendicular to the rule. [1] Ensuring for correct position of screen: Place the screen at the 50 .0 cm mark of the rule and move it slowly towards the lenses. Move the screen a little bit away and towards the lenses to ensure correct position of the screen. [1] Ensuring for focused image: Stop when you see that the image of the triangle has sharp corners. 2b [1] v = answer to 1 dp with units in the range of (0.133 – 0.147) [1] Averaging 2-3 values to increase accuracy. v = (13.9 + 14.1) / 2 = 14.0 cm (±5%) 2ci 1 𝑓 = 1 𝑢 + 1 𝑣 1 𝑓 = 𝑣 𝑢𝑣 + 𝑢 𝑢𝑣 1 𝑓 = 𝑣+𝑢 𝑢𝑣 [1] 𝑢𝑣 = 𝑓 (𝑢 + 𝑣) (PROVEN)

2cii 1. Assemble the apparatus as shown in Fig. 2.1. Follow the steps before (a) and in 2(a). 2. Determine an accurate value for the distance v by averaging two values [1] (or more) of v. 3. Record your values of u and v in a table. Include columns for uv and (u + v). For six values of u from 15.0 cm to 25.0 cm, determine corresponding values of v, uv and (u + v). [1] Record all of your measurements and calculations in your table. __u__ cm __v__ cm __uv__ cm2 __u + v__ cm 15.0 17.0 19.0 21.0 23.0 25.0 4. Plot a graph of uv against (u + v). [1] 5. The gradient [1] of your line of best fit is numerically equal to the focal length f of the combination of lenses. Determine f. Aim: T o determine the focal length of the combination of lenses IV: object distance u DV: image distance v CV: - (frequency of) light from torch - temperature/humidity of air - material (refractive index) of lenses - shape of lenses [Total: 10] 3ai [1] V = answer to 0.5 ml, with unit, in the range of (59.0 – 61.0) Read to half of the smallest division of the measuring cylinder. V = 60.0 ml (±1.5%) Twarm = 30.0°C 3aii [1] Correct calculation of mw with unit and correct s.f mw = 1.0 x 60.0 (2sf x 3sf) = 60 g (2sf) 3b [1] Tfinal = answer to 0.5°C, with unit Read to the half of the smallest division of the thermometer. 3c [1] Correct headings with quantities and units (as shown below) [1] Range of volumes evenly distributed (60, 70, 80, 90, 100, 110) [1] Correct trend (as V increases, Tfinal increases) [1] Correct dp for measurements and correct sf for calculations [1] Correct calculations and rounding 𝑉 𝑐𝑚3 𝑚𝑤 𝑔 1 𝑚𝑤 𝑔−1 𝑇𝑓𝑖𝑛𝑎𝑙 °C 1 𝑇𝑓𝑖𝑛𝑎𝑙 °C−1 to 1 dp to 2 sf to 2 sf to 1 dp to 3 sf

3di [1] axes labelled with quantities and units [1] good scale (not odd, points span > 50% of graph paper) [1] points correctly plotted [1] best fit line drawn 3dii Correct G calculated from the student’s graph. [1] Correct working (gradient triangle and 2 pairs of coordinates) [1] Correct answer, correct unit and correct sf (according to student’s graph) 3diii Correct cb calculated from the student’s G. [1] Correct working [1] Correct answer, correct unit and correct sf G = 𝑐𝑏 × 𝑚𝑏 𝑐𝑤 × 𝑇𝑤𝑎𝑟𝑚 cb = 𝐺 × 𝑐𝑤 × 𝑇𝑤𝑎𝑟𝑚 𝑚𝑏 = 0.303 × 4.2 × 30.0 100 = 0.38178 ≈ 0.38 J (g °C)-1 3div [1] Correct Y obtained/calculated from the student’s graph. (Correct working) Correct answer, correct unit and correct sf Y = 0.033 °C-1 1 𝑌 = 30 °C [1] Compares values of 1 𝑌 from (div) and Twarm in (a), and explains that the (percentage) difference is insignificant/significant. Yes, agree with suggestion since value of Twarm from (a) is 30.0°C and there is no difference between 1 𝑌 in (div) and Twarm in (a). 3e [1] SOE and [1] Improvement: There is energy transferred from the surroundings to the styrofoam cup and its contents, hence affecting the accuracies of Tfinal. Use a lid to reduce the energy transfer. The temperature of the tap water may not stay constant, hence affecting the accuracies of Twarm and Tfinal. Measure the temperature of the tap water using data loggers. The starting temperatures of the styrofoam cup may not have enough time to reset in between readings, hence affecting the accuracies of Twarm, and hence Tfinal. Use more sets of cups, so as to allow time for reset. Accept: Heat is gained during transfer between the ice bath and the Styrofoam cup. Reduce the temperature of the room using air conditioning/temperature controlled enclosure. [Total: 10]

𝑉 𝑐𝑚3 𝑚𝑤 𝑔 1 𝑚𝑤 𝑔−1 𝑇𝑓𝑖𝑛𝑎𝑙 °C 1 𝑇𝑓𝑖𝑛𝑎𝑙 °C−1 to 1 dp to 2 sf to 2 sf to 1 dp to 3 sf 50.0 50 0.020 25.4 = 25.0 0.0394 55.0 55 0.018 25.7 = 26.0 0.0389 60.0 60 0.017 26.0 0.0385 65.0 65 0.015 26.2 = 26.0 0.0381 70.0 70 0.014 26.5 = 27.0 0.0378 75.0 75 0.013 26.7 = 27.0 0.0375 80.0 80 0.013 26.8 = 27.0 0.0373 There is energy transferred from the surroundings to the brass, hence affecting the accuracies of Tcold and Tfinal. Use a larger mass of brass to reduce the rise in temperature of the brass when it is moved from the water bath to the styrofoam cup.