Assessed mathematical skills and measurement conventions

Treat every unit as part of the calculation. Substitute units with the numerical values, simplify powers and quotients, and check that the final unit matches the physical quantity being calculated.

Convert to compatible units before substitution. For example, $5.5\,\mathrm{mA}=5.5\times10^{-3}\,\mathrm{A}$, so a current calculated in amperes can be compared fairly with the measurement.

When a length is squared or cubed, its conversion factor is also powered: $1\,\mathrm{cm^3}=(10^{-2}\,\mathrm{m})^3=10^{-6}\,\mathrm{m^3}$.

Standard form writes a number as $a\times10^n$, where $1\leq |a|<10$ and $n$ is an integer. It preserves significant figures while making very large and very small physical quantities easier to combine.

Use physical constants with their stated precision and units. For light, $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$. A time of $2.20\,\mathrm{\mu s}$ is $2.20\times10^{-6}\,\mathrm{s}$.

At $0.980c$, the distance travelled in $2.20\,\mathrm{\mu s}$ is $s=0.980(3.00\times10^8)(2.20\times10^{-6})=6.47\times10^2\,\mathrm{m}$.

A ratio compares quantities in the same unit; a fraction expresses one quantity relative to a whole or reference; multiplying that fraction by $100\%$ gives a percentage. State the denominator because it defines the comparison.

If a device receives $240\,\mathrm{J}$ and transfers $180\,\mathrm{J}$ usefully, $\eta=180/240=0.75=75\%$. The remaining fraction is not automatically all one named loss unless the energy pathways are known.

A reciprocal is also a fraction: if $\lambda=5.00\times10^{-7}\,\mathrm{m}$, then $1/\lambda=2.00\times10^6\,\mathrm{m^{-1}}$. Transform the unit as well as the number.

An estimate predicts magnitude or direction of change from the governing relationship before detailed calculation. Round inputs sensibly, keep the dominant powers and use proportional scaling to expose impossible results.

For radiation intensity $I\propto1/d^2$, doubling distance gives $I_2/I_1=(d_1/d_2)^2=(1/2)^2=1/4$. This estimate predicts the direction and factor without knowing the luminosity.

A change that increases the measured value while leaving a similar absolute reading uncertainty reduces percentage uncertainty. This can justify changing an experimental parameter, provided the model and apparatus remain valid.

Use brackets and the calculator power key to evaluate transformed quantities such as $1/d^2$ or $x^{3/2}$. A negative exponent means a reciprocal; a fractional exponent means a root combined with a power.

For exponential decay, the ratio $N/N_0$ is dimensionless and lies between 0 and 1, so its natural logarithm is negative; the leading minus sign then gives a positive time.

Enter the complete exponent in brackets and verify whether the problem requires $\ln$ or $\log_{10}$. Substitute the result back into the original expression as a check.

Match calculator mode to the angle unit: degrees for angles marked $^\circ$, radians for angles stated in rad or produced by angular relationships. The same numerical input gives different results in the two modes.

For resultant components $R_x=4.0\,\mathrm{N}$ and $R_y=3.0\,\mathrm{N}$, the direction is $\theta=\tan^{-1}(3.0/4.0)=36.9^\circ$ above the positive $x$ direction.

The inverse-tangent value alone can be ambiguous. Use the signs of both components—or a calculator's two-argument angle function when available—to place the vector in the correct quadrant and state the reference direction.

Replace the prefix by its power of ten, then convert. For example, $3.6\,\mathrm{MW}=3.6\times10^6\,\mathrm{W}$ and $420\,\mathrm{nm}=420\times10^{-9}\,\mathrm{m}=4.20\times10^{-7}\,\mathrm{m}$.

For powered units, power the factor: $1\,\mathrm{cm^2}=10^{-4}\,\mathrm{m^2}$ and $1\,\mathrm{cm^3}=10^{-6}\,\mathrm{m^3}$.

Prefix symbols are case-sensitive: $\mathrm{M}$ means mega while $\mathrm{m}$ means milli and also serves as the unprefixed symbol for metre depending on position. Keep the quantity unit visible when interpreting the symbol.

Significant figures communicate measurement precision. A calculated result must not claim finer precision than the least accurate measurement used to obtain it.

A rectangle measured as $2.4\,\mathrm{cm}$ by $3.68\,\mathrm{cm}$ has calculator area $8.832\,\mathrm{cm^2}$. The $2.4\,\mathrm{cm}$ reading is quoted to 2 significant figures, so report $A=8.8\,\mathrm{cm^2}$, not $8.832\,\mathrm{cm^2}$.

Leading zeros are not significant, but zeros between non-zero digits and stated trailing zeros are: $0.00450$ has 3 significant figures. Scientific notation makes the intended precision explicit: $4.50\times10^{-3}$.

For $n$ repeated readings of the same quantity, the arithmetic mean is $\bar{x}=(x_1+x_2+\cdots+x_n)/n$. It reduces the influence of random variation but does not remove a systematic error.

For times $5.23$, $5.89$, $5.66$ and $5.01\,\mathrm{s}$, $\bar{t}=(5.23+5.89+5.66+5.01)/4=5.4475\,\mathrm{s}$, reported as $5.45\,\mathrm{s}$.

If each reading is for several cycles, first average the repeated total times and then divide by the number of cycles. For example, a mean time of $7.55\,\mathrm{s}$ for $5T$ gives $T=7.55/5=1.51\,\mathrm{s}$.

A probability lies from 0 to 1. In radioactive decay it describes the chance that a nucleus decays during a stated interval; identical nuclei have the same chance, but the time at which any one nucleus decays is unpredictable.

If each of $500$ nuclei has probability $0.020$ of decaying in a short interval, the expected number of decays is $Np=500\times0.020=10$. A particular observation need not contain exactly 10 decays.

A larger sample gives a more stable fraction of decays, while individual counts still fluctuate randomly. Probability predicts the behaviour of many repeated trials or a large population, not a deterministic result for one trial.

An order of magnitude is a power-of-ten scale. Write each variable as a coefficient times $10^n$, perform the exponent arithmetic separately, then normalise the result to identify its scale.

If $y=ab/c$, with $a\approx3\times10^6$, $b\approx2\times10^{-4}$ and $c\approx5\times10^2$, then $y\approx(6/5)\times10^{6-4-2}=1.2\times10^0$. Its order is $10^0$.

Round inputs enough to reveal the scale, but keep the coefficient while combining terms because it may shift the normalised power of ten. Compare the estimate with the calculator result to catch an incorrect exponent or prefix.

For the simple worst-case treatment used here, first express each measurement with an absolute or fractional uncertainty. The operation that combines the measured values determines how their uncertainties combine.

If $L_1=(12.4\pm0.1)\,\mathrm{cm}$ and $L_2=(8.2\pm0.1)\,\mathrm{cm}$, then $L_1+L_2=20.6\,\mathrm{cm}$ and $\Delta L=0.1+0.1=0.2\,\mathrm{cm}$. Report $(20.6\pm0.2)\,\mathrm{cm}$.

For $Q=A^2B/C$, $\Delta Q/Q=2(\Delta A/A)+(\Delta B/B)+(\Delta C/C)$. After adding the fractional terms, multiply by the calculated $Q$ to convert back to an absolute uncertainty when required.

A mathematical symbol states how quantities are related. Reading the symbol precisely prevents an approximation, inequality or proportionality from being treated as an exact equality.

This expression says that force increases with the rate of change of momentum. The two delta symbols define changes over the same interval; proportionality alone can be written as $F=k\Delta p/\Delta t$ only after introducing a constant $k$.

The subject is the quantity isolated on one side of an equation. Change it by applying inverse operations to both sides, preserving equality at every line.

To make $m$ the subject of $E=mc^2$, divide both sides by $c^2$: $E/c^2=m$, so $m=E/c^2$. The square applies to $c$ only, so it remains with the whole denominator.

Check the rearrangement by substituting it back into the original equation: $mc^2=(E/c^2)c^2=E$. A dimensional check also helps: $\mathrm{J}/(\mathrm{m^2\,s^{-2}})=\mathrm{kg}$.

Substitution replaces each symbol by its measured numerical value and unit. Convert quantities to a compatible unit system first, then keep the equation's structure visible during the calculation.

For $m=0.450\,\mathrm{kg}$ and $v=12.0\,\mathrm{m\,s^{-1}}$, $p=(0.450)(12.0)=5.40\,\mathrm{kg\,m\,s^{-1}}$. If mass were supplied as $450\,\mathrm{g}$, it would first become $0.450\,\mathrm{kg}$.

Solving a physics equation means finding every mathematical value of the unknown and then deciding which values satisfy the physical conditions. Kinematic equations such as $v=u+at$ and $s=ut+ frac12at^2$ apply only to constant acceleration.

If $s=20\,\mathrm{m}$, $u=2.0\,\mathrm{m\,s^{-1}}$ and $a=3.0\,\mathrm{m\,s^{-2}}$, then $20=2t+1.5t^2$, so $1.5t^2+2t-20=0$. The roots are about $3.05\,\mathrm{s}$ and $-4.38\,\mathrm{s}$.

For elapsed time after the stated start, the positive root is the relevant solution, so report about $3.0\,\mathrm{s}$ at suitable precision. Substitution into the original equation checks both the arithmetic and the chosen root.

A logarithm turns a ratio spanning many orders of magnitude into a compact scale value. On a base-10 logarithmic scale, adding 1 to the logarithm means multiplying the original ratio by 10.

The reference $Q_0$ makes the logarithm's argument dimensionless. If $Q/Q_0=10^3$, then $S=3$; if the ratio becomes $10^5$, the scale rises by 2 even though the original quantity is multiplied by $10^2=100$.

Sound intensity level is a real-world example: $L=10\log_{10}(I/I_0)\,\mathrm{dB}$. An intensity ratio of $100$ gives $L=10\log_{10}(100)=20\,\mathrm{dB}$ relative to $I_0$.

Numerical, algebraic and graphical forms can describe the same physical relationship. Translation means preserving the variables, units and conditions while changing how the relationship is represented.

In a linear stress-strain region with stress on the vertical axis and strain on the horizontal axis, Young modulus $E$ is the gradient. A stress of $120\,\mathrm{MPa}$ at strain $6.0\times10^{-4}$ gives $E=120\times10^6/(6.0\times10^{-4})=2.0\times10^{11}\,\mathrm{Pa}$.

A useful graph gives each measured pair an unambiguous position and uses the plotting area efficiently. For ‘extension against force’, plot extension vertically and force horizontally.

If uncertainty bars are supplied or required, draw them to the stated uncertainty in the correct direction. A best-fit line should balance the overall scatter rather than be forced through the origin or through every point.

Before interpreting the plot, verify that scale increments are uniform, every point lies within the axes, and transformed variables such as $1/x$ or $\ln x$ are labelled as the quantities actually plotted.

A relationship is linear in the plotted variables when it can be written as $y=mx+c$, where $m$ and $c$ are constants. Identifying $y$, $x$, $m$ and $c$ predicts the graph before it is drawn.

A curved relationship may become linear after a justified transformation. The transformed quantity—not the original symbol alone—must occupy the axis used in the comparison with $y=mx+c$.

For constant acceleration, $v=u+at$ predicts a straight velocity-time graph: acceleration fixes its gradient and initial velocity fixes its vertical intercept.

For a best-fit straight line, choose two well-separated points on the line; they need not be measured data points. Use a large gradient triangle, retain the sign and obtain gradient units from vertical-axis unit divided by horizontal-axis unit.

The vertical intercept is the value of $y$ when $x=0$. On a velocity-time graph it can represent initial velocity. If the plotted axis is transformed, reverse that transformation: an intercept $log_{10}A_0=1.57$ gives $A_0=10^{1.57}$.

A line through $(1.0\,\mathrm{s},5.0\,\mathrm{m\,s^{-1}})$ and $(7.0\,\mathrm{s},17.0\,\mathrm{m\,s^{-1}})$ has gradient $12.0/6.0=2.0\,\mathrm{m\,s^{-2}}$.

When a graph is linear, its gradient gives one constant rate of change across the whole interval. The physical rate follows from the quantities and units on the axes.

If a straight velocity-time line rises from $4.0\,\mathrm{m\,s^{-1}}$ at $t=1.0\,\mathrm{s}$ to $16.0\,\mathrm{m\,s^{-1}}$ at $t=7.0\,\mathrm{s}$, $a=(16.0-4.0)/(7.0-1.0)=2.0\,\mathrm{m\,s^{-2}}$.

A horizontal line has zero rate. A negative gradient gives a negative rate relative to the chosen positive direction; for velocity this is negative acceleration, not automatically a decrease in speed.

For a curved graph, the rate changes from point to point. The gradient of a tangent at the chosen point estimates the instantaneous rate there because the tangent matches the curve's local direction.

On a displacement-time graph, suppose a tangent at $t=2.0\,\mathrm{s}$ passes through convenient tangent points $(1.0\,\mathrm{s},3.0\,\mathrm{m})$ and $(3.0\,\mathrm{s},11.0\,\mathrm{m})$. The instantaneous velocity is $(11.0-3.0)/(3.0-1.0)=4.0\,\mathrm{m\,s^{-1}}$.

A longer tangent triangle reduces the percentage effect of reading uncertainty. The two calculation points belong to the tangent and need not lie on the original curve.

An average rate describes change across a finite interval; an instantaneous rate describes the rate at one particular point. On a curved graph they are generally different.

If displacement changes from $2\,\mathrm{m}$ at $1\,\mathrm{s}$ to $14\,\mathrm{m}$ at $5\,\mathrm{s}$, average velocity is $(14-2)/(5-1)=3\,\mathrm{m\,s^{-1}}$. A tangent at $5\,\mathrm{s}$ could have a different gradient.

As the interval around a point becomes smaller, its chord gradient can approach the tangent gradient when the curve is smooth. This explains why a tangent represents the local rate without requiring explicit differentiation.

The area between a curve and the horizontal axis can represent a physical quantity when multiplying the axis units produces that quantity. Its meaning must come from the model, not from geometry alone.

For straight sections, add rectangle, triangle or trapezium areas. For a curve, estimate with narrow strips or count squares. Treat area below the horizontal axis as negative when the represented quantity is signed.

For a linear capacitor voltage-charge graph rising from zero to $12\,\mathrm{V}$ at $4.0\,\mathrm{mC}$, the triangular area is $\tfrac12(12)(4.0\times10^{-3})=2.4\times10^{-2}\,\mathrm{J}$. This application is full A Level content.

A rate equation links a quantity's present value to how quickly it changes. It can be explored with graph gradients or a spreadsheet using small finite time steps, without writing derivatives or integrals.

With $x=10$, $\lambda=0.20\,\mathrm{s^{-1}}$ and $\Delta t=1.0\,\mathrm{s}$, the initial rate is $-2.0$ units $\mathrm{s^{-1}}$ and the next modelled value is $10+(-2.0)(1.0)=8.0$. Repeating the rows produces a decaying curve.

For capacitor discharge $V=V_0e^{-t/\tau}$, taking a logarithm makes voltage linear in time. The gradient reveals the time constant $\tau$, but its formula depends on the logarithm used.

If a graph of $\ln(V/V_{\mathrm{ref}})$ against $t$ has gradient $-0.250\,\mathrm{s^{-1}}$, then $\tau=-1/(-0.250)=4.00\,\mathrm{s}$.

Changing the voltage reference or stated voltage unit shifts the intercept but not the gradient, provided one consistent convention is used for every point.

A proposed exponential or power law can be tested by transforming it into a straight-line form. The transformed graph must be linear and its gradient must agree with the proposed parameter.

Radioactive decay and capacitor discharge give negative gradients on $\ln y$ against time. For $F=kx^{-2}$, a log-log graph should be straight with gradient $-2$; a measured gradient far from $-2$ does not support the inverse-square claim.

Use several transformed data points, appropriate axes and a best-fit line. Judge agreement using scatter and measurement uncertainty rather than demanding an exact textbook gradient from imperfect data.

A sketch shows qualitative shape, intercepts, turning behaviour and asymptotes implied by an equation. For positive $k$, the function family predicts the following features before any numerical scale is chosen.

Physical domains can retain only part of the mathematical graph: pressure and volume are positive, time may begin at zero, and a squared physical quantity may not use the negative-$x$ branch.

Changing a positive constant $k$ stretches the vertical scale without changing the function family. A negative $k$ reflects the graph across the horizontal axis.

Angles in a physical diagram come from the geometry of the structure and from the direction in which each vector acts. Establish these angle relationships before choosing a calculation.

On an inclined plane, weight remains vertically downward, the normal contact force is perpendicular to the plane, and friction is parallel to the plane. The plane angle therefore fixes the complementary angles used when resolving weight.

In a regular three-dimensional structure, identify which edges or axes are genuinely perpendicular or parallel. A perspective drawing may make a right angle look oblique, so use stated geometry rather than apparent page angle.

A two-dimensional representation selects the plane and directions needed to solve a three-dimensional physical situation. A force diagram then isolates one object and shows only the external forces acting on it.

For a block on a rough incline, show weight vertically downward, normal contact force perpendicular to the surface and friction parallel to the surface opposing actual or impending relative motion.

A 2D projection can omit a third coordinate only when no required force or displacement component lies outside the chosen plane. Otherwise use separate perpendicular components or another view.

Choose a geometry formula that matches the actual shape, convert every length to one unit system, and power the conversion factor with the dimension of the result.

For a wire of diameter $0.400\,\mathrm{mm}$, $A=\pi(0.400\times10^{-3})^2/4=1.26\times10^{-7}\,\mathrm{m^2}$. With $L=2.00\,\mathrm{m}$ and $\rho=1.70\times10^{-8}\,\Omega\,\mathrm{m}$, $R=0.270\,\Omega$.

Pythagoras' theorem relates only the sides of a right-angled triangle. Perpendicular vector components form such a triangle, so their resultant magnitude is the hypotenuse.

For components $R_x=6.0\,\mathrm{N}$ east and $R_y=8.0\,\mathrm{N}$ north, $R=\sqrt{6.0^2+8.0^2}=10.0\,\mathrm{N}$. The direction is found separately from the component triangle.

A triangle's interior angles total $180^\circ$. To check whether measured sides $a$, $b$ and longest side $c$ make a right angle, test whether $a^2+b^2=c^2$ within measurement uncertainty.

Sine, cosine and tangent connect a vector to a right-triangle representation. First identify the angle's reference axis; the adjacent component uses cosine and the opposite component uses sine.

A $50.0\,\mathrm{N}$ force at $30.0^\circ$ above the horizontal has $F_x=50.0\cos30.0^\circ=43.3\,\mathrm{N}$ and $F_y=50.0\sin30.0^\circ=25.0\,\mathrm{N}$.

For a resultant with components $R_x$ and $R_y$, $\tan\theta=R_y/R_x$. Use the signs of both components to choose the correct quadrant and state the direction relative to an axis.

For a sufficiently small angle measured in radians, $\sin\theta\approx\theta$, $\tan\theta\approx\theta$ and $\cos\theta\approx1$. These approximations replace a curved trigonometric relationship by a simple linear one.

For wavelength $\lambda=600\,\mathrm{nm}$, screen distance $D=2.00\,\mathrm{m}$ and slit separation $s=0.500\,\mathrm{mm}$, the fringe spacing is $w=(600\times10^{-9})(2.00)/(0.500\times10^{-3})=2.40\,\mathrm{mm}$.

One radian is the central angle that subtends an arc equal in length to the radius. This definition makes angular relationships such as arc length and phase naturally dimensionless.

A phase difference of $25.0^\circ$ is $25.0\pi/180=0.436\,\mathrm{rad}$. Conversely, $1.20\,\mathrm{rad}=1.20(180/\pi)=68.8^\circ$.