Physics 9702/52 May/June 2024: question paper and mark scheme

Mark scheme

1 Defining the problem t is the independent variable and s is the dependent variable or vary t and measure s 1 keep k constant 1 Methods of data collection labelled diagram of workable experiment including: 1 spring connected to magnet vertical rule parallel to spring to determine s rule held in position by a stand stand resting on the bench rule labelled and at least one other label from stand, clamp, card, (magnetic) sheet, (cylindrical) magnet, spring s = (new) length/position of spring – original length/position of spring 1 use a micrometer to measure t 1 measure B using a (calibrated) Hall probe and rotate probe until maximum value 1 or measure B using Hall probe first in one direction, then in the opposite direction and average 1 Method of Analysis 1 1 plot a graph of s against or equivalent t (allow lg s against lg t) relationship valid if a straight line that passes through the origin is produced 1 (for lg s against lg t: relationship valid if a straight line with gradient −1) k 1 Z= × gradient ALB k × 10 y -intercept (for lg s against lg t: Z = ) ALB 1 Additional detail including safety considerations 6 D1 precaution related to spring and/or magnet hitting eyes, e.g. use of goggles/use of safety screen around experiment D2 keep A, L and B constant D3 use a rule to measure L D4 micrometer/calipers to measure diameter d of the magnet and A = πd2 / 4 D5 description of method to determine k, e.g. add mass to spring and k = mg / extension or use newton meter to measure force applied to spring and k = force / extension or take several readings of force and extension, plot a force–extension graph and k = gradient D6 (magnetic) sheet clamped to bench D7 use pointer(s)/marker(s) on the spring to read off values from the rule D8 method to use video recorder and replay to determine maximum length of the spring or increase s or force gradually/slowly until magnet (just) leaves the card D9 repeat measurements of t in different positions on the card and average t or repeat measurements of s for each value of t and average s D10 method to check that the spring has not exceeded the elastic limit D11 use of non-magnetic stand or named non-magnetic material for stand, e.g. wood

Mark scheme

2(a) gradient = n 1 2 y-intercept = lg C 2(b) 1 lg (L / cm) lg (T / 10–5 s) 1.73 or 1.732 1.38 or 1.380 ± 0.02 1.85 or 1.845 1.51 or 1.505 ± 0.01 1.93 or 1.934 1.59 or 1.591 ± 0.01 2.033 or 2.0334 1.69 or 1.690 ± 0.02 2.146 or 2.1461 1.81 or 1.806 ± 0.01 2.223 or 2.2227 1.87 or 1.869 ± 0.01 Values of lg (L / cm) and lg (T / 10–5 s) correct as shown above. Uncertainties in lg (T / 10–5 s) correct as shown above. 1 2(c)(i) Six points from (b) plotted correctly. 1 Must be within half a small square. Diameter of points must be less than half a small square. Error bars in lg T plotted correctly. 1 All error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical. 2(c)(ii) Straight line of best fit drawn. 1 Do not accept line from top point to bottom point. Line must pass between (1.780, 1.45) and (1.800, 1.45) and between (2.085, 1.75) and (2.100, 1.75) Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1 All error bars must be plotted. 2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx. 1 Distance between data points must be greater than half the length of the drawn line. Gradient determined of worst acceptable line with clear substitution of data points into Δy / Δx. 1 uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) 2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1 y-intercept of worst acceptable line determined by substitution into y = mx + c. 1 uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line or uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept) Do not accept ECF from false origin method. 2(d) Value of n determined using gradient (n = gradient) and C given to 2 or 3 significant figures. 1 Value of C determined using y-intercept with method shown. 1 2 C= y -intercept 10 Absolute uncertainties in n and C. 1 uncertainty in n = uncertainty in gradient and 2 2 − 2 min worst y -intercept max worst y -intercept ΔC = C − or ΔC = 10 10 10 worst y-intercept 2 Clear method must be shown with ΔC correctly evaluated. 2(e) Value of L determined (non-zero) to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d) with correct substitution 1 and correct power of ten. Units of T and either C or y-intercept must be consistent. 2 logT − log log L = C = log10 − y -intercept n n log10 − y -intercept L = 10 n or TC L=n 2