RVHS 2026 J2 Prelims P4 MS

1 2026 JC2 Prelims P4 Qns Marking Instructions Mark 1(a) V0 recorded to the nearest 0.01 V. 1 1(b)(i) value of t = 12 s to 17 s evidence of repeated readings of t 1 1(b)(ii) accept Δt from 0.2 to 0.6 s percentage uncertainty of t ≈ 2.5% percentage uncertainty expressed to 1 s.f. or 2 s.f. 1 1(c)(i) 6 sets of data without assistance and table headings correct, with units shown (table heading should include t, I1, I2, Imean, ln Imean) repeated readings of I for each t 1 All raw data recorded and expressed to correct precision (I to nearest 0.01 mA or 0.001 mA) 1 ln Imean calculated correctly and d.p. corresponding to s.f. of Imean 1 Comments: Many students expressed the precision of ln Imean incorrectly. Many students recorded the value of current incorrectly, such as recording it as 6.02 mA instead of 0.602 mA. A small number of students did not repeat readings for I. 1(c)(ii) working showing how B and A can be determined from gradient and y-intercept respectively 1 sketch of suitable graph with axes labelled AND trendline equation shown 1 A and B determined correctly, and units shown (unit of A = A or mA, unit of B = s) 1 Comments: Many students labelled the unit for A incorrectly. It is important to note that in an equation t BAe − =I , the unit for A follows the unit for I. A small number of students fitted an exponential curve instead of linearising it. The equation has to be linearised because the question referred to a trendline. By convention, ‘line’ refers to a straight line. Plotting the straight line also allows us to see and assess the fit more clearly, as we can visually check for linearity. 1(d)(i) The absolute uncertainty in t remains the same because it is caused by human reaction time. As the rate of change of I decreases, the absolute uncertainty in I also decreases. 1

2 1(d)(ii) The total amount of charge transferred can be determined by taking the area under the I-t graph. 1 Comments: This question was generally done well. Students who did not score the mark commonly failed to state clearly which graph should be plotted, specifically identifying the quantities to be plotted on the y-axis and x-axis. 1(d)(iii) amount of charge transferred ≈ 0.0150 C 1 Comments: • A number of students gave the final answer with incorrect units. Students should take care to note that the readings in the table are typically recorded in mA and, hence, the estimated charge should be expressed in mC. • A number of students obtained estimates that differed significantly from the expected value. Inspection of the working suggested that this was due to incorrect use of Excel. In particular, some students used a formula to calculate the area of each trapezium and then dragged the formula down for the subsequent rows, but inadvertently included the final row, which should have been excluded. Note that if there are 6 data points, there should only be 5 trapeziums. Example of mistake made in Excel: These are not valid data. Excel assumes that its value is 0 and continues calculation without providing an error message. There should not be a 6th trapezium. Recommended: Start from the second row, so you will not forget about it at the last row. Total: 12 marks ✔

3 Qns Marking Instructions Mark 2(a) t = 0.770 mm Accept value of t in range of 0.730 – 0.800 mm 1 Comments: While repeated thicknesses were not marked for this time, it is good practice to actually do so to ensure that there are no anomalies. Some candidates took measurements at the wrong end and their readings were skewed by the thickness of the tape used for labelling. (b)(i) T1 = 0.6910 s • total timing of oscillations NT to be more than 20 s • Number of oscillations N should be clearly stated • Accept value of period T1 in range of 0.60 to 0.80 s • Take repeated readings of total timings NT. • NT to the nearest 0.01 s with correct units. • T to follow s.f. of NT. 1 Comments: Most were able to get credit for this but there are still some candidates who presented their answers in 2 s.f, not realising that the raw data was in 4 s.f. (ii) 1f T= = 1.447 Hz f calculated corrected and presented with correct s.f. and units. 1 (iii) 1 f T tf T f T t   =  = = Percentage uncertainty in f calculated corrected using appropriate uncertainty for T Δt between 0.2 to 0.5 s 1 Comments: Candidates should take note to present their workings clear. There were some who calculated the percentage uncertainty using the frequency or period directly, it is important that they show clearly how the corresponding absolute uncertainty was obtained (iv) T2 less than T1. NT to the nearest 0.01 s with correct units. 1 (c)(i) 25.4 cm Sum up amplitudes of crests and troughs from t = 0 to t = 2 s Accept value within (25.2 – 25.5 cm), units included 1 Comments: Quite a number of candidates misread the question and seemed to have calculated the total distance travelled for the entire dataset.

4 Qns Marking Instructions Mark (ii) Maximum speed = 20.3 cm s−1 Accept value with 10% difference (17.9 – 22.8 cm s−1), units included 1 Comments: Most students were able to get this, small number did not include the proper units. (iii) Explanation corresponding to use of steepest gradient for first oscillation. 1 Comments: Most candidates were able to give an satisfactory explanation. (iv) sketch of negative -sloping graph with axes labelled AND trendline equation shown 1 Gradient of graph is approximately -0.36 if at least 6 pts is selected (accept – 0.45 to -0.35) 1 Comments: This part proved challenging to some candidates. They were not able to get an equation for the trendline. Some also misread the question and simply drew the variation with time of the displacement of the particle. y = -20.274x + 5.5288 R² = 0.9997-3.00 -2.00 -1.00 0.00 1.00 2.00 3.00 0.00 0.10 0.20 0.30 0.40 0.50 Chart Title y = -0.377x + 4.5667 R² = 0.985 0.00 1.00 2.00 3.00 4.00 5.00 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 Chart Title

5 Qns Marking Instructions Mark c) measurement of displacement and corresponding time using an appropriate method for fast oscillations, with a suitable means of measuring/calibrating displacement and appropriate positioning of the measuring device (e.g. video analysis with ruler/scale in the plane of motion; motion sensor fixed relative to the laboratory and positioned such that the blade moves towards/away from the sensor, connected to a data logger to obtain displacement–time data)) 1 method to determine amplitude from displacement–time data by identifying the maximum displacements from equilibrium position for each oscillation (e.g. identify successive peak values for each oscillation/ turning points) 1 plot of ln(amplitude) against time to test for exponential relationship. If a straight line/ constant gradient (negative) is obtained, relationship is valid. 1 reference to valid improvement method to improve identification of maxima (e.g. use a higher frame-rate recording so that the turning point can be located more precisely / interpolate between adjacent frames near the maximum) control of conditions to ensure exponential damping (small amplitude / constant damping conditions) method to reduce uncertainty in amplitude measurement (parallax-free / perpendicular recording / fixed scale) secure clamping of blade to prevent injury from oscillating blade / secure with blu-tack 1 Comments: This planning was challenging for the candidates. For measuring the displacement and corresponding time, some candidates simply mentioned to use camera, not realising that while the camera can be used to show the motion of the fast-moving hacksaw blade, there will still need to be a means to show the actual displacem