| Date | May 2011 | Marks available | 2 | Reference code | 11M.2.SL.TZ2.1 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | State | Question number | 1 | Adapted from | N/A |
Question
Data analysis question.
The speed v of waves on the surface of deep water depends only on the wavelength λ of the waves. The data gathered from a particular region of the Atlantic Ocean are plotted below.
The uncertainty in the speed v is ±0.30 m s–1 and the uncertainty in λ is too small to be shown on the diagram.
State, with reference to the graph,
(i) why v is not directly proportional to λ.
(ii) the value of v for λ=39m.
It is suggested that the relationship between v and λ is of the form
\[v = a\sqrt \lambda \]
where a is a constant. To test the validity of this hypothesis, values of v2 against λ are plotted below.
(i) Use your answer to (a)(ii) to show that the absolute uncertainty in v2 for a wavelength of 39 m is ±5 m2 s–2.
(ii) The absolute uncertainty in v2 for a wavelength of 2.5 m is ± 1m2 s–2. Using this value and the value in (b)(i), construct error bars for v2 at the data points for λ=2.5 m and 39 m.
(iii) State why the plotted data in (b)(ii) suggest that it is likely that v is proportional to \(\sqrt \lambda \).
(iv) Use the graph opposite to determine the constant a.
(v) Theory shows that \(a = \sqrt {\frac{k}{{2\pi }}} \). Determine a value for k.
Markscheme
(i) the graph is not linear/a straight line (going through the error bars) / does not go through origin [1]
(ii) 7.7 ms−1 ; (N.B. line is drawn for candidate, answer must be correct) [1]
(i) % uncertainty in \(v = \left( {\frac{{0.3}}{{7.7}} = } \right)3.9\% \);
doubles 3.9% (allow ECF from (a)(ii)) to obtain % uncertainty in \({v^2}\)(=7.8%);
absolute uncertainty (=±[0.078×59.3])=4.6;
(=±5m2s−2)
or
calculates overall range of possible value as 7.4−8.0; (allow ECF)
squares values to yield range for v2 of 54.8 to 64; (allow ECF)
so error range becomes 9.2 hence ±4.6; (must see this value to 2 sig fig or better to award this mark)
(ii) correct error bars added to first point (±½ square) and last-but-one point (±2.5 squares); (judge by eye)
(iii) a straight-line/linear graph can be drawn that goes through origin;
(iv) uses triangle to evaluate gradient; { (triangle need not be shown if read-offs clear, read-offs used must lie on candidate’s drawn line) to arrive at gradient value of 1.5±0.2; (unit not required)
recognizes that gradient of graph is a2 and evaluates a=1.2±0.2(m½s−1);
or
candidate line drawn through origin and one data point read;
correct substitution into v2=a2λ; (a2 does not need to be evaluated for full credit)
a=1.2±0.2(m½s−1);
Award [2 max] if line does not go through origin – allow ½ square.
Award [1 max] if one or two data points used and no line drawn.
(v) k=9.4ms−2; (allow ECF from (b)(iv))
Examiners report
(i) Most were able to identify one (of several reasons) why the proportionality did not apply.
(ii) Almost all could state the value at the required point to a sensible accuracy.
(i) Many fully understood the simple treatment of combination of errors and arrived at a correct and well-explained solution.
(ii) The error bars were usually correctly drawn, however in a small number of cases, candidates drew the same length bar for both points (usually using the value for the upper data point).
(iii) Unlike in (b) the reasons for proportionality were usually incomplete on this occasion and few candidates scored the mark. The fact that the line goes through the origin was often ignored.
(iv) This question was done poorly; the work of many candidates was very disappointing here. Only about half the candidates attempted to draw a straight line on the graph (they were told to “Use the graph”) and simply used two points on the graph without reference to a line. This gained little credit as the candidate gave no evidence at all that the chosen pair of points both lay on the line. Candidates then often compounded this by quoting a2 as the answer to the question, failing to recognize that a square root was required.
(v) Most candidates were able to take their derived a (correct or not) and evaluate k however the unit of k was usually ignored.
