Resource for Students and Scientists

Dr John Cornish

Physics and Energy Studies

Division of Science and Engineering

This presentation describes the process of problem solving in physics. The approach adopted here involves the use of a general problem solving algorithm.

- To describe an algorithm for solving Physics problems
- To apply the algorithm to examples

- Step 1: recognise the problem
- Step 2: interpret the information given
- Step 3: recall relevant facts, retrieve data, that may be useful but were not given
- Step 4: focus on the problem
- Step 5: work through
- Step 6: check back

Ask yourself:

- Have I seen a problem like this before?
- Is it a problem with a verbal description?
- Are some data given?
- Is a calculation required to obtain a numerical answer?
- Is a verbal response required?

- Carefully read any verbal description and restate it in your own words
- It is usually appropriate to draw a diagram
- Add label and dimensions
- Accurately copy figures and formulas given to your notebook

Write down any additional information that might be useful:

- Data
- Equations

At this point you should have before you all the information necessary to achieve a solution.

- You should now review this information.
- Identify and discard items that now appear redundant.
- Do not be too vigorous with crossing out, you may need this information if your first attempt at a solution is not successful.

- Is a calculation required?
- Check that there are as many equations as there are variables
- Make a guess at the form or magnitude of the answer
- Decide on the first moves towards the solution

If a verbal answer is required, write your first draft.

If a numerical answer is required:

- Solve the appropriate equation(s)
- Evaluate the answer(s) numerically
- Remember units!

If a graphical answer is required:

- Draw an appropriate graph.
- Add a title and label the axes.
- Make sure the calculated or given points are clearly indicated.
- Draw an appropriate best straight line or curve through the points.

You now have an answer but should ask yourself; Is it physically possible?

- Is the magnitude reasonable?
- Are the units correct?
- Are there spelling or grammatical errors?
- Are there any algebraic errors?
- Are there any numerical errors?

If the answers to the previous questions are satisfactory, then the answer to the problem is complete.

If errors are obvious, then make appropriate corrections.

If you are uncertain, particularly in an examination, be cautious about making changes;

Many a correct answer has been crossed-out and replaced by an incorrect one.

Two types of seismic waves (P and S waves) travel at 6.23 and 3.58 km/s respectively through the Earth’s crust in South Australia.

A person in Adelaide feels two jolts, separated by 4 seconds, corresponding to the arrival of these waves.

How far away is the source of the earthquake?

This is a problem that looks similar to many thunderstorm problems where there are two speeds involved.

Unlike the thunderstorm problems where the speed of light is sufficiently large that it can be taken as instantaneous, both seismic wave speeds are similar and we might expect that two equations will be required.

These will need to be solved simultaneously.

We have two uniform velocities:

\begin{equation*} v_1 = 6.23 ~\text{km/s} \text{ and } v_2 = 3.58 ~\text{km/s} \end{equation*}We know that the waves will travel the unknown distance $s$ in times $t_1$ and $t_2$ respectively such that

\begin{equation*} t_1 - t_2 = 4 ~\text{seconds} \end{equation*}Recall the relationship between distance time and uniform velocity:

\begin{equation*} v = \dfrac{s}{t} \end{equation*}Set up the three equations that describe the situation:

\begin{equation*} v_1 = \dfrac{s}{t_1}\text{, } v_2 = \dfrac{s}{t_2} \text{ and } t_1 - t_2 = 4 \end{equation*}We now have three equations and three unknowns and a solution is possible.

Notice that we do not need to evaluate $t_1$ and $t_2$ as these were not asked for in the problem.

We can now either:

- (a) solve the equations algebraically and then put in numbers, or
- (b) put in the numbers now and then solve the equations.

Method (a) is more general and useful if there are several sets of figures to be substituted in the equations, but (b) is acceptable if a “one off” solution is required.

For practice we shall follow method (a).

The first two equations may be rearranged to give:

\begin{equation*} t_1 = \dfrac{s}{v_1}\text{, } t_2 = \dfrac{s}{v_2} \end{equation*} Substituting for $t_1$ and $t_2$ in the third equation gives \begin{equation*} \dfrac{s}{v_1} - \dfrac{s}{v_2} = 4 \end{equation*}or

\begin{equation*} s \left(\dfrac{1}{v_1} - \dfrac{1}{v_2}\right) = 4 \end{equation*}Leading to

\begin{equation*} s = \dfrac{4}{\left(\dfrac{1}{v_1} - \dfrac{1}{v_2}\right)} \end{equation*}Expressing the equation as;

\begin{equation*} s = 4 \times \dfrac{1}{\left(\dfrac{1}{v_1} - \dfrac{1}{v_2}\right)} \end{equation*}and putting in numbers:

\begin{equation*} s = 4 \times \dfrac{1}{\left(\dfrac{1}{3.58} - \dfrac{1}{6.23}\right)} \text{ km} \end{equation*}In this form and using the reciprocal key on your calculator will simplify the calculation to get

\begin{equation*} s = 4 \times 8.416 \text{ km} \end{equation*}leading to $s$ = 33.67 km.

There are several ways to estimate the magnitude.

If we assume that the faster wave arrives instantaneously, then the distance the second wave travels in 4 seconds would be 4 × 3.58 km or 14.4 km.

Since we haven’t taken the finite speed of the first wave into account, this will give an underestimate of the distance, so we would expect the actual distance to be somewhat greater than 14.4 km.

To get an upper limit on the distance is not so easy. We could say, for example, that since in the absence of any other information, we have assumed a “flat Earth”, then the distance is probably much less then the radius of the Earth.

\begin{equation*} r_E = 6.37 \times 10^3 \text{ km} \end{equation*}Since our answer is greater than 14.4 km and certainly less than 6.37 × 10^{3}
km we may have some confidence in its correctness.

We have simply assumed that given the speed in km/s and the time in seconds, that the distance will automatically come out in km. To check,

\begin{equation*} s = \dfrac{v}{t} \end{equation*}with units of (km/s)/s. The seconds cancel out and we are left with km as expected.