Physics 9702/51 May/June 2019: question paper and mark scheme

Mark scheme

1 Defining the problem L is the independent variable and s is the dependent variable or vary L and measure s 1 keep mass of load or M constant 1 Methods of data collection labelled diagram of workable experiment including: 1 • method of fixing strip at one end, e.g. with a G-clamp or (heavy) mass placed on top of strip over bench • load shown touching at point P • load and G-clamp or (heavy) mass labelled method of attaching load to strip, e.g. use glue/tape/attach with a hook and string 1 use a rule to measure L and s 1 use a balance to measure M 1 Method of analysis plot a graph of s against L3 (allow lg s against lg L, or log) 1 relationship valid if a straight line through (0,0) 1 (for lg s against lg L gradient of straight line = 3) 4Mg 1 E= 3 bt × gradient 4Mg (for lg s against lg L, E = 3 ÷ 10 y −intercept ) bt 1 Additional detail including safety considerations Max. 6 D1 use cushion/foam/sandbox in case mass/load falls or wear goggles in case strip snaps or recoils D2 use same wooden strip or keep b and t constant D3 clamp rule vertically to measure s D4 method to ensure clamped rule to measure s is vertical, e.g. correctly positioned set square indicated at right angles to the horizontal surface or plumb line shown in appropriate position D5 s = reading of vertical rule with loaded strip – reading of vertical rule with no load D6 repeat s measurement for each L (unloading and loading strip) and average s D7 use a micrometer/calipers to determine t D8 repeat readings for b and/or t at different points along/across the strip and average D9 method to ensure strip is perpendicular to the bench, e.g. repeat measurements of L on each side of strip to check that L is constant or set square correctly indicated on diagram D10 wait until block is stationary/in equilibrium or measure s after a fixed time

Mark scheme

2(a) K 1 gradient = E 2(b) 1 1 –1 /A I 180 or 175 210 or 213 250 290 or 286 320 or 323 370 1 1 uncertainties in from ±3 or ±4 to ±10–15 I 2(c)(i) Six points plotted correctly. 1 Must be accurate to the nearest half 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. Length of bar must be accurate to less than half a small square and symmetrical. 2(c)(ii) Line of best fit drawn. 1 Do not allow line from top point to bottom point. If points are plotted correctly then lower end of line should pass between (122, 230) and (126, 230) and upper end of line should pass between (186, 350) and (190, 350). 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 points from the line into ∆y/∆x. 1 Distance between points must be at least half the length of the drawn line. uncertainty = gradient of line of best fit – gradient of worst acceptable line 1 or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) 2(d) 9.4 ± 0.2 (V) 1 2(e)(i) K determined from gradient and given to 2 or 3 significant figures. 1 K = E × gradient = 9.4 × (c)(iii). K determined from gradient with correct unit (Ω / °). 1 2(e)(ii)   ∆gradient   ∆E   1 % uncertainty in K =   +   × 100   gradient   E   2(f) θ calculated. 1 Correct substitution of numbers required. E 9.4 θ= = I K 0.01× (e)(i) or 1 1 θ= = I × gradient 0.01× (c)(iii) Absolute uncertainty in θ. 1 Correct substitution of numbers required. Use of ∆I not required but allow if included by the candidate. Using E and K:  ∆E ∆K  ∆I   uncertainty inθ =  + +  ×θ  E K  I    0.2 (e)(ii)  uncertainty inθ =  +  ×θ  9.4 100  max E min E max θ = or minθ = 0.01× min K 0.01× max K Using gradient:  ∆gradient  ∆I   uncertainty inθ =   +  ×θ  gradient  I   1 1 max θ = or minθ = 0.01× min gradient 0.01× max gradient