Eduqas A-Level Physics

SI base units

Every physical quantity is measured in a unit that can be built from a small set of base units. Six of them are needed for this course.

Every other unit is a derived unit, a product or quotient of base units. The newton is the force that gives a mass of 1 kg an acceleration of $1\,\mathrm{m\,s^{-2}}$, so $1\,\mathrm{N} = 1\,\mathrm{kg\,m\,s^{-2}}$. The joule is a newton metre, so $1\,\mathrm{J} = 1\,\mathrm{kg\,m^2\,s^{-2}}$, and the watt is a joule per second, so $1\,\mathrm{W} = 1\,\mathrm{kg\,m^2\,s^{-3}}$. The coulomb is an ampere second, the volt is a joule per coulomb, which gives $1\,\mathrm{V} = 1\,\mathrm{kg\,m^2\,s^{-3}\,A^{-1}}$, and the ohm is a volt per ampere, so $1\,\Omega = 1\,\mathrm{kg\,m^2\,s^{-3}\,A^{-2}}$.

Prefixes

A prefix multiplies the unit by a power of ten. The ones used most are tera (T, $10^{12}$), giga (G, $10^9$), mega (M, $10^6$), kilo (k, $10^3$), centi (c, $10^{-2}$), milli (m, $10^{-3}$), micro ($\mu$, $10^{-6}$), nano (n, $10^{-9}$), pico (p, $10^{-12}$) and femto (f, $10^{-15}$). The prefix belongs to the whole unit that follows it, so $1\,\mathrm{cm^3} = (10^{-2}\,\mathrm{m})^3 = 10^{-6}\,\mathrm{m^3}$, not $10^{-2}\,\mathrm{m^3}$. This is the commonest source of a wrong power of ten when converting areas and volumes.

Checking an equation for homogeneity

An equation is homogeneous when every term has the same units. A correct physical equation must be homogeneous, so writing each side in base units is a check on an equation that has been remembered or derived. It cannot prove an equation right, because it is blind to pure numbers such as $\tfrac12$ or $2\pi$, but it will catch a wrong power or a missing quantity.

As an example, the equation for the period of a simple pendulum is $T = 2\pi\sqrt{l/g}$. The units of $l/g$ are $\mathrm{m}/\mathrm{m\,s^{-2}} = \mathrm{s^2}$, so the square root has the unit of seconds, which matches $T$. The same idea can be used to find an unknown unit. The Young modulus is stress over strain, strain has no unit, and stress is $\mathrm{N\,m^{-2}}$, so the Young modulus is in $\mathrm{N\,m^{-2}} = \mathrm{kg\,m^{-1}\,s^{-2}}$.

Scalars and vectors

A scalar quantity has magnitude only. A vector quantity has magnitude and direction. The distinction matters because vectors must be added with their directions taken into account.

A vector is drawn as an arrow whose length represents its magnitude, pointing in its direction. Two opposite directions along a line are distinguished by a sign, and the sign is part of the vector: a velocity of $-3\,\mathrm{m\,s^{-1}}$ is three metres per second in the direction taken as negative.

Adding vectors

Two vectors are added by placing the tail of the second at the head of the first. The resultant goes from the tail of the first to the head of the second. The order does not matter. Subtracting a vector is the same as adding its reverse, and a reversed vector has the same length and the opposite direction. Only two perpendicular vectors need be combined by calculation. If $A$ and $B$ are at right angles, the resultant $R$ has

where $\theta$ is the angle between $R$ and $A$. A boat heading due north at $4.0\,\mathrm{m\,s^{-1}}$ through water in a river that flows east at $3.0\,\mathrm{m\,s^{-1}}$ has a velocity relative to the bank of magnitude $\sqrt{4.0^2 + 3.0^2} = 5.0\,\mathrm{m\,s^{-1}}$, at $\tan^{-1}(3.0/4.0) = 37^\circ$ east of north.

Resolving a vector

Any vector can be replaced by two perpendicular vectors, its components, which together have the same effect. A vector of magnitude $V$ at angle $\theta$ to the horizontal has components

so the component along the direction of the angle is the cosine and the one opposite the angle is the sine. Which is which depends on where the angle is measured from, and it is worth checking the extreme cases: at $\theta = 0$ the whole vector is horizontal, and the cosine is 1.

Resolving is what makes inclined planes and projectiles tractable. A trolley of weight $W = 12\,\mathrm{N}$ on a slope at $25^\circ$ to the horizontal has a component of weight down the slope of $W\sin25^\circ = 12\times0.4226 = 5.1\,\mathrm{N}$ and a component into the slope of $W\cos25^\circ = 10.9\,\mathrm{N}$. The second is balanced by the normal contact force of the slope.

Density

The density of a material is its mass per unit volume.

$\rho$ (rho) is in $\mathrm{kg\,m^{-3}}$, $m$ in kg and $V$ in $\mathrm{m^3}$. Water is about $1000\,\mathrm{kg\,m^{-3}}$, which is $1.00\,\mathrm{g\,cm^{-3}}$, and since $1\,\mathrm{g\,cm^{-3}} = 10^{3}\,\mathrm{kg\,m^{-3}}$ the conversion between the two is a factor of 1000.

For a uniform object the density is that of the material. A steel sphere of radius $1.5\,\mathrm{cm}$ and mass $0.109\,\mathrm{kg}$ has a volume $\tfrac43\pi r^3 = \tfrac43\pi(0.015)^3 = 1.41\times10^{-5}\,\mathrm{m^3}$ and a density of $0.109/1.41\times10^{-5} = 7.7\times10^{3}\,\mathrm{kg\,m^{-3}}$.

The volumes needed are $\pi r^2 h$ for a cylinder, $lwh$ for a rectangular block and $\tfrac43\pi r^3$ for a sphere.

Measuring the density of a solid

For a regular solid, measure the mass on a balance and the dimensions with instruments matched to their size: a metre rule for lengths of tens of centimetres, vernier callipers for lengths of a few centimetres and a micrometer screw gauge for small diameters such as a wire. Repeat each length in several places and take a mean, because real objects are not perfectly uniform, and check the zero reading of each instrument first, since a zero error is a systematic error that does not average out.

The volume is worked out from the mean dimensions, and the percentage uncertainty in the density is the sum of the percentage uncertainty in the mass and the percentage uncertainties in each length, counting a length twice if it is squared and three times if it is cubed. So the smallest dimension is usually the one that limits the precision: a thin rod's diameter enters as $d^2$, and a small absolute error in it matters much more than one in its length.

For an irregular solid, the volume can be found by displacement. Lower the object into a measuring cylinder of water and read the rise in level, or use a displacement can and collect the water that overflows. The object must be fully submerged, bubbles on its surface removed, and the scale read at eye level at the bottom of the meniscus.