Paper 3 Measurement Answers

Question 1: Measurement model, transformation and graph analysis [15]

(a) The model predicts $T^2$ is directly proportional to $L$ [1]. It therefore predicts a straight line through the origin on a graph of $T^2$ against $L$ [1].

(b)

$L/\mathrm{m}$	0.250	0.350	0.450	0.550	0.650
$T^2/{\mathrm{s}}^2$	1.00	1.42	1.80	2.19	2.59

Correct squaring method [1]; values [1]; sensible significant figures [1].

(c) Award independently:

• axes labelled with quantity and unit, with $T^2$ on the vertical axis [1];
• simple scales using at least half the available grid in both directions [1];
• all five points plotted accurately as small crosses [1];
• one straight best-fit line representing the overall trend, not a dot-to-dot join [1].

An indicative completed graph is shown below. Small differences in best-fit placement are acceptable.

(d) Use two well-separated points on the candidate’s best-fit line, not necessarily two measured points. For example, an acceptable line gives

$$k\approx 4.0{\mathrm{s}}^2{\mathrm{m}}^{-1}.$$

Large-triangle gradient method [1]; value consistent with the drawn line, typically $3.9$ to $4.1$ [1]; unit ${\mathrm{s}}^2{\mathrm{m}}^{-1}$ [1]. Apply follow-through from the candidate’s line.

(e) The total timed interval is larger [1], so a similar reaction-time uncertainty is a smaller percentage of the measured interval and hence of each period [1].

(f) Accept one relevant feature [1], for example a zero offset in the measured length, measuring length from the wrong reference point, or a timing-system delay that does not cancel between starting and stopping. Do not award merely “reaction time” without explaining a consistent offset; similar start and stop delays may cancel in an elapsed-time measurement.

Question 2: Longer structured vector investigation [20]

(a)

$$x_2=24.0\cos 60.0^{\circ }=12.0\mathrm{m},$$ $$y_2=24.0\sin 60.0^{\circ }=20.8\mathrm{m}.$$

Correct trigonometric components [2]; units/signs [1].

(b)

$$x=18.0+12.0=30.0\mathrm{m},y=20.8\mathrm{m}.$$

East component [1]; north component [1].

(c)

$$R=\sqrt{30.0^2+20.8^2}=36.5\mathrm{m}.$$ $$\theta ={\tan }^{-1}\left(\frac{20.8}{30.0}\right)=34.7^{\circ }$$

north of east. Magnitude method/value [2]; direction method/value [2].

(d) The two displacements are not along the same line [1], so direction matters and only components or a vector diagram can combine them correctly [1].

(e)

$$\frac{0.4}{24.0}\times 100\%=1.7\%.$$

Fractional or percentage method [1]; value [1].

(f) Changing the angle rotates the vector [1]. Since $x=R\cos \theta$ and $y=R\sin \theta$ [1], both component values depend on $\theta$ even if $R$ is fixed [1].

(g) Use a stated scale [1], draw the first vector east and the second vector head-to-tail at $60^{\circ }$ north of east [1], then measure the resultant from the starting point to final point with magnitude and direction [1].

(h) Scale drawings are limited by drawing thickness, protractor/ruler resolution, and human reading uncertainty [1].

Review Figures

Figure: Paper 3 graph reasoning normally expects correctly transformed quantities, clear axes, sensible units and physical interpretation of gradient or intercept. In Question 1, plotting $T^2$ against $L$ makes the proportional model testable.

Figure: The vector workflow in Question 2 is the component version of head-to-tail addition: resolve each displacement, add signed components, then reconstruct the resultant magnitude and direction.