(i) On the graph opposite, draw error bars on the first and last points to show the uncertainty in v.

(ii) On the graph opposite, draw the line of best-fit for the data points.

Explain whether the data support the hypothesis that v is directly proportional to W.

Theory suggests that the relation between v and W is

$$v = k{W^3}$$

where k is a constant.

To test this hypothesis a graph of ${v^{\frac{1}{3}}}$ against W is plotted.

At W=5.5 N the speed is 250±20 μm s^–1.

Calculate the uncertainty in ${v^{\frac{1}{3}}}$ for a load of 5.5 N.

(i) Using the graph in (c), determine k without its uncertainty.

(ii) State an appropriate unit for your answer to (d)(i).

(i) both error bars correct (overall length 4 squares) $\pm \frac{1}{2}$ square;

(ii) smooth curve going through error bars and within half square of other points;

not proportional because not straight/trend cannot go through origin;

fractional error in $v = \frac{{20}}{{250}}\left( { = 0.080} \right)$;
fractional error in ${v^{\frac{1}{3}}} = \frac{{0.080}}{3}\left( { = 0.027} \right)$; (allow ECF from first marking point)
uncertainty in ${v^{\frac{1}{3}}} = \left( {0.063 \times 0.027 = } \right)0.00169$; (allow 0.00168−0.00170)
Allow expression of answer as 0.630±0.002 if calculation above seen.
Award [3] for a bald correct answer.

or

recognizes uncertainty in ${v^{\frac{1}{3}}} = \frac{{\sqrt[3]{{270}} - \sqrt[3]{{230}}}}{2}$ or ${\sqrt[3]{{250}} - \sqrt[3]{{230}}}$ or ${\sqrt[3]{{270}} - \sqrt[3]{{250}}}$;
= 0.168 ;
conversion to 0.00168ms⁻¹;

(i) large triangle > half line used;
read-offs and substitution correct; (allow power of ten error here)
${k^{\frac{1}{3}}} = 0.012 \pm 0.001$; (allow ECF)
k=1.73×10^–6 m N^–3 s^–1; (allow correct power of ten only)
Award [0] for use of a single data point.

(ii) m N^–3 s^–1 or kg⁻³ m⁻² s⁵;