Physics 9702/51 October/November 2024: question paper and mark scheme

Mark scheme

1 Defining the problem s is the independent variable and v is the dependent variable or vary s and measure v 1 keep D constant 1 Methods of data collection labelled diagram of workable experiment including: 1 • light gate positioned at P • light gate connected to timer / data logger • labels for light gate and P and data logger / timer and at least one other label from block, magnet(s), trolley, s and D measure D with a rule(r) and measure L with a rule(r) or calipers 1 description to determine v at P, e.g. (measure length of) card to interrupt beam 1 method to measure s, e.g. use calipers 1 1 Method of Analysis 1 1 1 plot a graph of v2 against 4 or equivalent (e.g. 4 against v2) s s Do not accept logarithms. m × gradient 1 K= 2DA2B 2L2 m 1 (or K = for 4 against v2) 2DA B L × gradient 2 2 2 s m × y -intercept 1 Q=− 2D m × y -intercept 1 (or Q = KA2B 2L2 × y -intercept or Q = for 4 against v2) 2D × gradient s 1 Additional detail including safety considerations 6 D1 method to stop the trolley (after passing point P), e.g. labelled block / buffer / cushion drawn after P or place a block / buffer / cushion after P to stop the trolley D2 keep L, A, m and B constant πd 2 D3 use micrometer / calipers to measure diameter (d) of the magnet and A = 4 D4 method to secure block to bench, e.g. clamp block to bench or (heavy) mass on top of block or method to secure magnets, e.g. use glue to stick magnets to trolley / block D5 method to increase the accuracy of measuring s or D, e.g. use a marker to left of the trolley D6 measure B using a (calibrated) Hall probe and adjust / rotate probe until maximum value or measure B using Hall probe first in one direction, then in the opposite direction and average D7 use a (top-pan) balance to measure m D8 use of strong magnets to increase v D9 repeat measurements of v for each value of s and average v D10 relationship valid if a straight line is produced (passing through − 2DQ ) m Do not accept line passing through the origin.

Mark scheme

2(a) gradient = 3 1 3E − E s y-intercept = 2Z 3E − E s 2(b) 1 1 / A−1 I 5150 or 5155 5560 or 5556 5810 or 5814 6250 6670 or 6667 6940 or 6944 Values correct as shown above. 1 1 Uncertainties in from ±50 or ±60 to ±90 or ±100. I 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. 1 1 Error bars in plotted correctly. I 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. Points must be balanced. Line must pass between (1.63, 5400) and (1.67, 5400) and between (2.58, 6800) and (2.62, 6800). 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)(i) E determined using gradient 1 and E and Z given to 2 or 3 or 4 significant figures. 1 3 3 + gradient × Es 1 E E= + Es = = + s 3 gradient 3 × gradient gradient 3 1 E= + 0.733 gradient Z determined using y-intercept 1 and E and Z given with SI units with correct powers of ten. Z= ( 3E − Es ) × y -intercept or 3 × y -intercept Z= 2 2 × gradient Unit of E: V Unit of Z: Ω 2(d)(ii) Absolute uncertainty in E with method shown. 1 Δgradient 1 0.05 uncertainty = × + gradient gradient 3 or correct substitution for max/min methods. 2(e) Value of R determined to a minimum of two significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and 1 correct use of power of ten. 1 −6 − y -intercept R= 250 × 10 gradient or 1 2Z R= − gradient × 250 × 10 −6 3 or 3E − 2.2 2Z R= − 3 × 250 × 10−6 3