Exam technique and the problem-solving method

Builds on: Section S1 and school physics
Reading time: 45 min
Exercises: 2
Olympiad problems: 3

One short problem in the theory paper of IOAA 2009 asked how small a planet would have to be for a person to escape it by jumping. It gives no numbers at all, apart from saying that the planet is as dense as Earth. A student who knows every formula in the syllabus can still be stuck on a problem like this, because the difficulty lies in deciding what the problem is about, which numbers to supply and how to check the result. To earn marks, a student has to turn this knowledge into a written solution within the time allowed. In this section we set out the habits that do this, from sharing out the five hours to a method for attacking a problem that seems to give too little to work with.

S2.1Time and marks

The theory paper is long, and the time for each problem has to be planned. The marks for each problem are printed on the paper, so the first job is to turn them into minutes. A paper of 250 marks in 300 minutes gives 1.2 minutes for each mark, so a 10-mark short problem is worth about 12 minutes and a 30-mark long one about 36 minutes. About a tenth of the paper goes on reading it at the start and checking at the end, so each problem gets a little less time than this.

At the start of the paper we spend ten minutes or so reading every problem, noting which ones look quick and which look long. Short problems are best done early, because each one is a few safe marks. A long problem should be started once its first parts are clear, since those are usually the easiest marks in it. A problem that is going badly should be left once it has used up its share of the time, and Section S2.3 says what to write down before moving on.

S2.2Partial credit

Each solution is marked against a scheme that splits the marks of a problem between its steps. A typical scheme gives marks for stating the right principle, for writing the correct equation in symbols, for a correct intermediate result and for the final answer with its unit. Markers usually also follow a wrong result into later parts, so a slip in one part does not lose the marks of every part after it, as long as the later working is right.

This changes how a solution should be written. The working has to be visible, because a marker can only give marks for steps that are on the page. A formula written in symbols before any numbers go in shows the marker the method even if the arithmetic goes wrong later. When a part cannot be finished, writing down the principle and the equation that would solve it still earns marks, while a blank earns none. A final answer that is boxed or underlined, with its unit, is easy to find and to mark.

S2.3A method for olympiad problems

Textbook exercises usually practise one formula at a time, and they give exactly the data that formula needs. Olympiad problems instead describe a situation, often a real one, and leave the choice of physics to the student. Some give more data than is needed, and some, like the jumping problem, give almost none. The five steps below take a student from the physics to a written solution.

We start by drawing the situation, with the bodies, the observer, the angles and the distances, and we label everything that the problem gives. Many positional astronomy problems are solved by the drawing alone.
Next we name the physics. Most problems turn on one or two principles, such as conservation of energy, the cosine rule on the celestial sphere or the inverse-square law, and saying which ones apply narrows the search.
Then we work backwards from what is asked. If the problem wants a radius, we write down an equation that contains the radius and ask which quantities in it are still unknown.
We use every piece of information. Each number and each phrase in a problem is usually there for a reason, and one that seems surprising often points to the idea that the problem is testing.
In a problem with several parts, we treat the earlier parts as hints. The setters often split a long problem so that each part supplies a result the next part needs.

When a problem still will not give way, there are some standard ways out. Writing down the relevant principle and equations earns marks and often shows the next step. A limiting case, such as an observer at the pole or a planet of zero mass, can show what the answer must look like. An estimate with round numbers gives a target to check against. If none of these helps within a few minutes, it is time to move on and come back later.

Example S.1Jumping off a planet

How large is the largest planet, as dense as Earth, from which a person could escape by jumping straight up? This was a 10-mark problem at IOAA 2009, which gave no other data.

Solution

We draw a person on a small round planet, leaving the ground at speed $v$ . The physics is conservation of energy, and the person escapes if $v$ is at least the escape speed of the planet, $v_{esc} = \sqrt{2 G M / R}$ . The problem gives no numbers, so we supply them. A good jump on Earth lifts the centre of mass by about 0.5 m, and energy conservation on Earth gives the take-off speed,

v = \sqrt{2 g h} = \sqrt{2 \times 9.81 \times 0.5} = 3.13 {m s}^{- 1} .

Next we use the one fact the problem does give, that the density is the same as Earth's. The mass of a planet of radius $R$ and density $ρ$ is $\frac{4}{3} π R^{3} ρ$ , so

v_{esc} = \sqrt{\frac{2 G}{R} \cdot \frac{4}{3} π R^{3} ρ} = R \sqrt{\frac{8 π G ρ}{3}} .

For a fixed density the escape speed is proportional to the radius. Earth's escape speed is 11 180 m/s for a radius of 6378 km, so the planet we want has

R = 6378 km \times \frac{3.13}{11 180} = 1.79 km .

AnswerThe planet could have a radius of about 1.8 km. As a check on the size of this answer, Deimos, the smaller moon of Mars, has a radius of about 6 km and an escape speed of about 5.6 m/s, so even there a jump would fall back.

The supplied number, the height of a jump, is an assumption, and a solution should say so. Markers accept any sensible value, such as 0.3 m or 0.6 m, as long as it is stated and used correctly. The answer changes only as the square root of the jump height, so a rough estimate is good enough.

S2.4Units, figures and assumed values

Answers need units, normally SI units, or the units the problem itself uses, such as parsecs, hours or arcseconds. Checking the units of a formula before putting numbers into it catches many mistakes, because both sides of a correct equation always have the same units. If $\sqrt{G M / R}$ comes out in metres per second, as it does, it can be a speed.

The answer should be given to about as many significant figures as the least precise of the data, usually two or three, so with data given to three figures a calculator display of 7.678723 is written as 7.68. Intermediate results, though, should be kept at full precision in the calculator's memory rather than copied down rounded and typed back in, because rounding at every step adds up to errors in the last figure, and sometimes in more than that.

When a problem needs a value that it does not give, we use the table of constants supplied with the paper rather than a value remembered from elsewhere, so that the answer agrees with the marking scheme. When a value is in neither, as with the jump height above, we choose a sensible one and state it clearly as an assumption.

S2.5Using the calculator

The IOAA allows only a non-programmable, non-graphing scientific calculator, and in 2026 every contestant was given the same model, a Casio fx-82CW. Practising with a simple scientific calculator of this kind, rather than with a phone or a graphing calculator, avoids surprises in the exam.

Before every trigonometric calculation we check the angle mode. Working out $\sin 30 °$ should give 0.5, and if it gives −0.988 the calculator is in radian mode. Angles in degrees, minutes and seconds go in with the calculator's sexagesimal key, which converts 23°26′ to 23.43° without a separate step. Intermediate results go into the calculator's memories and are reused from there, which keeps their full precision.

Example S.2A better jumperfinish it

A good athlete can lift their centre of mass by 0.8 m in a jump. How large a planet of Earth's density could the athlete escape from?

The start

The take-off speed is $v = \sqrt{2 \times 9.81 \times 0.8} = 3.96$ m/s. By Example S.1, the radius is proportional to the take-off speed.

Your stepNow find the radius in kilometres.

Common errorsSection S2

A final answer written without its unit loses marks, even when the number is right.
Rounding intermediate results and typing them back in adds up errors. Intermediate values belong in the calculator's memory at full precision.
A calculator left in radian mode gives wrong values for every angle in degrees, and the error is easy to miss. Working out $\sin 30 °$ first shows which mode is set.
An assumed value, such as the height of a jump, should be stated as an assumption. A marker cannot give credit for a number that seems to come from nowhere.
A student who stays too long on one stubborn part usually loses more marks than the part is worth. Writing the principle down, marking the part and coming back later is almost always better.

ExercisesSection S22 questions

S2.1
What is the unit of $\sqrt{G M / R}$ , where $G$ is in N m² kg⁻², $M$ in kg and $R$ in m?
S2.2
A person who can jump 0.5 m on Earth tries to escape from a planet as dense as Earth but with a radius of only 1 km. Assume the take-off speed is the same as on Earth. By how much, in metres per second, does the take-off speed exceed the planet's escape speed?
m/s

Olympiad problemsPast IOAA problems, restated3 problems

2009Jumping off a planet10 points

Adapted from IOAA 2009 (Tehran), theory problem 3. We have restated it in our own words and kept the original data. The solution was last checked on 16 October 2026.

Estimate the radius of a planet from which a person could escape by jumping straight up. Assume that the planet has the same density as Earth.

I solved it without the solution

AnswerThe planet could have a radius of about 1.8 km. Any sensible jump height is acceptable if it is stated, and the radius changes only as the square root of the height.

2010The longest life of the Sun10 points

Adapted from IOAA 2010 (Beijing), theory problem 5. We have restated it in our own words and kept the original data, adding the Sun's mass and luminosity from the IOAA 2025 data sheet. The solution was last checked on 16 October 2026.

If 0.8% of the Sun's initial mass could be turned into energy during its whole life, what is the longest the Sun could live, assuming that its luminosity stays the same? Take the Sun's mass as $1.988 \times 10^{30}$ kg and its luminosity as $3.828 \times 10^{26}$ W.

I solved it without the solution

AnswerThe Sun could shine for at most about $1.2 \times 10^{11}$ years. Its real life on the main sequence is about ten times shorter, because only the hydrogen in its core, about a tenth of its mass, is ever turned into helium (Chapter T7).

2011Solar neutrinos10 points

Adapted from IOAA 2011 (Katowice), theory problem 14. We have restated it in our own words and kept the original data, adding the solar constant from the IOAA 2025 data sheet. The solution was last checked on 16 October 2026.

Estimate the number of solar neutrinos that pass every second through an area of 1 m² of Earth's surface, held perpendicular to the Sun's rays. Each fusion reaction in the Sun releases 26.8 MeV of energy and two neutrinos. Take the solar constant, the power of sunlight on 1 m² at Earth's distance, as 1361 W m⁻², and 1 eV as $1.602 \times 10^{- 19}$ J.

I solved it without the solution

AnswerAbout $6 \times 10^{14}$ neutrinos pass through each square metre every second, by night as well as by day.