Edexcel GCSE Physics

SI units

Every physical quantity is measured in a unit, and physics uses the International System (SI) so that a value means the same thing everywhere. Six base units are needed at GCSE, and every other unit in the course is built from them.

Some combinations of base units are used so often that they have their own name. These are the derived units the course uses.

Temperature differences are often given in degrees Celsius, and a change of 1 °C is the same size as a change of 1 K. Units such as $\mathrm{m/s}$ and $\mathrm{kg/m^3}$ are compound units, made by dividing one unit by another, and the way they are written tells you how the quantity is built: speed is distance per unit time, and density is mass per unit volume.

A calculation gives a number and a unit together. An answer of 12 with no unit says nothing, and a wrong unit usually means a step has gone wrong, so it is worth checking that the unit of the answer is the one the quantity should have.

Multiples and sub-multiples

Values that are very large or very small are written with a prefix in front of the unit. The prefixes needed are these.

To convert to the base unit, multiply by the power of ten. A frequency of 2.5 GHz is $2.5 \times 10^{9}$ Hz, and a current of 470 mA is $470 \times 10^{-3} = 0.47$ A. To go the other way, divide by the power of ten. An energy of 3,600,000 J is $3\,600\,000 \div 10^{6} = 3.6$ MJ.

Two conversions cause most of the slips. Areas and volumes use the prefix twice or three times, so a square centimetre is $(10^{-2})^2 = 10^{-4}$ square metres, and 1 $\mathrm{cm^3}$ is $10^{-6}\ \mathrm{m^3}$. A block of volume $250\ \mathrm{cm^3}$ is therefore $250 \times 10^{-6} = 2.5 \times 10^{-4}\ \mathrm{m^3}$. The other is time, which is not counted in tens: there are 60 seconds in a minute and $60 \times 60 = 3600$ seconds in an hour.

Converting between units

Convert by multiplying or dividing by the ratio between the two units, and let the units guide the arrangement so that the one you want to remove cancels.

A vehicle travels at $72\ \mathrm{km/h}$. There are 1000 m in 1 km and 3600 s in 1 h, so

The same method turns hours into seconds. A lesson that lasts 1.5 hours is $1.5 \times 3600 = 5400$ s, and a 45 minute drive is $45 \times 60 = 2700$ s. Equations in physics are written for SI units, so a quantity given in kilometres, grams, minutes or millijoules should be converted before it goes into an equation, and the answer then comes out in the SI unit.

The same care applies to rates. A rate compares two quantities, and it can be worked out by proportional reasoning. If a machine transfers 900 J in 15 s, then it transfers $900 \div 15 = 60$ J each second, which is 60 W. Ratios are written with no unit when the two quantities have the same unit, as with the efficiency of a device or the magnification of a lens.

Significant figures and standard form

The significant figures of a number are the digits that carry information, counted from the first non-zero digit. So 3,450 has three significant figures if the final zero only shows the place of the digit, 0.0072 has two, and 0.4050 has four. To round to a number of significant figures, look at the digit after the last one you keep. If it is 5 or more, round up.

In a calculation, give the answer to the same number of significant figures as the least precise value used, or one more than that when it is a step on the way to a longer calculation. Rounding a value in the middle of a calculation and using the rounded value causes a small error in the answer, so keep the full calculator value until the end and round once.

Standard form writes a number as $A \times 10^{n}$, where $A$ is at least 1 and less than 10, and $n$ is a whole number. It is used for very large and very small numbers, since a physicist reads the size of a number from its power of ten.

Multiplying and dividing in standard form uses the rules of powers. Multiply the numbers, add the powers, and put the answer back into standard form.

If the number part comes out as 10 or more, or below 1, change the power to fix it. For example $(5 \times 10^{4}) \times (4 \times 10^{3}) = 20 \times 10^{7} = 2 \times 10^{8}$.

Orders of magnitude

To estimate, round each value to the nearest power of ten and compare. An atom has a diameter of the order of $10^{-10}$ m, and a nucleus is of the order of $10^{-15}$ m across, so the nucleus is about $10^{5}$ times smaller than the atom. Answers found this way are given with the approximation symbol $\approx$ or $\sim$, and the aim is to check that a calculated value is sensible before trusting it: a person walking at 30 m/s, or a car with a mass of 1.2 kg, cannot be right.