Cambridge International A-Level Physics

Physical quantities

A physical quantity is a property that can be measured, and it always has two parts: a numerical magnitude and a unit. A length of $2.4\,\mathrm{m}$ is not described by the $2.4$ alone, since the same number of centimetres or miles would be a different length. Every equation in physics relates quantities, so a calculation that loses its units has lost half of its meaning.

Some quantities also have a direction. Those are vectors, and they are treated in the note on vectors and describing motion.

SI base quantities

The SI system is built on a small set of base quantities, from which every other quantity is derived. The five used at this level are these.

A sixth base quantity, amount of substance, has the unit mole (mol) and is used with gases and the Avogadro constant.

Derived units

A derived unit is a product or quotient of base units. Every other unit is one of these, even when it has a name of its own. The work is to reduce a named unit to base units by going back to an equation that defines the quantity.

Velocity is displacement divided by time, so its unit is $\mathrm{m\,s^{-1}}$. Acceleration is velocity divided by time, giving $\mathrm{m\,s^{-2}}$. Force is mass times acceleration, so the newton is

Energy and work are force times distance, so the joule is

Power is energy divided by time, so the watt is

Pressure is force divided by area, so the pascal is

Charge is current times time, so the coulomb is $\mathrm{C} = \mathrm{A\,s}$. Potential difference is energy per unit charge, so the volt is

Resistance is potential difference divided by current, so the ohm is

Frequency has the hertz, $\mathrm{Hz} = \mathrm{s^{-1}}$. Density is mass per volume, in $\mathrm{kg\,m^{-3}}$. The same method works for any other quantity met later: write the defining equation and substitute the units of everything in it.

Homogeneity of equations

An equation is homogeneous when every term has the same base units. Terms that are added or subtracted must have the same units, and the two sides of an equals sign must match. This gives a quick test for whether an equation could be right. It cannot show that an equation is right, because numerical factors and dimensionless quantities such as $\pi$ have no units and so are invisible to the test.

To test $s = ut + \tfrac12 at^2$, take each term in base units. The left side is in $\mathrm{m}$. The term $ut$ is $\mathrm{m\,s^{-1}} \times \mathrm{s} = \mathrm{m}$. The term $\tfrac12 at^2$ is $\mathrm{m\,s^{-2}} \times \mathrm{s^2} = \mathrm{m}$. All three agree, so the equation is homogeneous.

The same test finds the units of a constant. The period of a pendulum is thought to follow $T = k\sqrt{l/g}$. The units of $\sqrt{l/g}$ are $\sqrt{\mathrm{m}/\mathrm{m\,s^{-2}}} = \mathrm{s}$, which matches $T$, so $k$ has no unit. If an equation were proposed as $T = k\sqrt{lg}$, the right side would have units $\sqrt{\mathrm{m^2\,s^{-2}}} = \mathrm{m\,s^{-1}}$ for $\sqrt{lg}$, which cannot equal a time unless $k$ carried units of $\mathrm{s^2\,m^{-1}}$, so the form is rejected.

A calculation that ends in the wrong units has gone wrong somewhere, which makes checking the units of a final answer worthwhile.

Prefixes

Prefixes scale a unit by a power of ten so that very large and very small quantities can be written compactly.

The prefix belongs to the unit and is raised to the power with it. So $1\,\mathrm{cm^2}$ is $(10^{-2}\,\mathrm{m})^2 = 10^{-4}\,\mathrm{m^2}$, not $10^{-2}\,\mathrm{m^2}$, and $1\,\mathrm{mm^3} = 10^{-9}\,\mathrm{m^3}$. Conversions done in the wrong direction are a common source of errors of a factor of a thousand or a million.

Worked example

A wire has a diameter of $0.36\,\mathrm{mm}$. Its cross-sectional area in square metres is

A frequency of $2.4\,\mathrm{GHz}$ is $2.4\times10^{9}\,\mathrm{Hz}$, and a wavelength of $650\,\mathrm{nm}$ is $6.5\times10^{-7}\,\mathrm{m}$.

Estimating physical quantities

An estimate is a value that is right to within about a factor of ten, reached by reasoning from sizes and quantities known well enough. Estimates are used to check that a calculated answer is sensible and to judge which effects matter. A few values are worth knowing at this level:

To estimate the kinetic energy of a cyclist, take a mass of $80\,\mathrm{kg}$ for rider and bicycle and a speed of $6\,\mathrm{m\,s^{-1}}$. Then $E_\mathrm{K} = \tfrac12 mv^2 = \tfrac12 \times 80 \times 36 \approx 1.4\times10^{3}\,\mathrm{J}$, and it is stated as roughly a kilojoule, not to three figures. The precision of an estimate should match the precision of the inputs.