Paper 2 Measurement Answers

Question 1: Units, prefixes and homogeneity [10]

(a) $L$: $\mathrm{m}$ [1]; $g$: $\mathrm{m}{\mathrm{s}}^{-2}$ [1].

(b)

$$\frac{L}{g}=\frac{\mathrm{m}}{\mathrm{m}{\mathrm{s}}^{-2}}={\mathrm{s}}^2$$

Correct substitution of units [1]; correct simplification [1].

(c) $\sqrt{L/g}$ has unit $\mathrm{s}$ [1]; $2\pi$ is dimensionless, so the right-hand side has unit $\mathrm{s}$, matching $T$ [1].

(d) Dimensional analysis cannot detect a wrong dimensionless numerical factor [1] or whether the physical model/assumptions are appropriate [1].

(e)

$$75.0\mathrm{cm}=0.750\mathrm{m}=7.50\times 10^{-1}\mathrm{m}.$$

Correct conversion [1]; correct standard form with sensible significant figures [1].

Question 2: Uncertainty in a derived quantity [12]

(a)

$$V=lwh=(5.20)(3.10)(1.60)=25.792\mathrm{c}{\mathrm{m}}^3\approx 25.8\mathrm{c}{\mathrm{m}}^3.$$

Method [1]; value/unit [1].

(b)

$$\rho =\frac{84.0}{25.792}=3.2576\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}\approx 3.26\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}.$$

Method [1]; value/unit [1].

(c)

$$\frac{\Delta V}{V}\times 100\%=\frac{0.05}{5.20}\times 100\%+\frac{0.05}{3.10}\times 100\%+\frac{0.05}{1.60}\times 100\%=5.70\%.$$

Product-rule use [1]; all three percentage terms [1]; correct result [1].

(d) For $\rho =m/V$, percentage uncertainties add:

$$\frac{\Delta \rho }{\rho }\times 100\%=\frac{0.1}{84.0}\times 100\%+5.70\%=5.82\%.$$ $$\Delta \rho =0.0582(3.2576)=0.190\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}.$$

Mass uncertainty included [1]; correct total percentage [1]; absolute uncertainty [1].

(e)

$$\rho =3.26\pm 0.19\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$$

or $3.3\pm 0.2\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ with consistent rounding. Value and uncertainty consistent [1]; unit included [1].

Figure: The density calculation uses multiplication/division, so fractional or percentage uncertainties are added. The exact constants in the rectangular-volume expression do not contribute additional uncertainty.

Question 3: Vector addition and data presentation [11]

(a) Displacement has magnitude and direction [1].

(b)

$$R=\sqrt{120^2+90^2}=150\mathrm{m}.$$

Pythagoras [1]; value/unit [1].

(c)

$$\theta ={\tan }^{-1}\left(\frac{90}{120}\right)=36.9^{\circ }.$$

Correct ratio [1]; direction $36.9^{\circ }$ north of east [1].

(d) Components after the westward motion:

$$x=120-50=70\mathrm{m},y=90\mathrm{m}.$$

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

(e) Distance is the scalar path length travelled [1], while displacement is the vector change from start to finish and depends only on final position relative to start [1].

(f) Any two: arrows drawn head-to-tail; clear scale or stated not-to-scale; directions labelled; resultant drawn from start to final position; components/axes labelled [2].

Figure: A clear vector diagram places displacement vectors head-to-tail and draws the resultant from the first tail to the final head. This is why the final displacement is not obtained by simply adding path lengths.