Dimensional Analysis
(Aside: whenever I write these posts, the default title reads "Write an engaging title", then I go an title the post "Dimensional Analysis"... oh well, I suppose the people interested in these sorts of things probably find that exciting...)
What's written here looks kinda complicated, but in fact it's one of the laziest ways to solve a problem possible - and that's why it's useful.
One of my Oxford interview questions ran like this: "A ball is travelling with speed v when it rolls into a well of depth h and width w. If the ball bounces back and forth before landing on the ground, how many times will it bounce?"
This sounds complicated. One approach is to find how far it's fallen after the first bounce, then after the second and so on. This approach is very long and probably doesn't even work (how do you know when you've gone a height larger than h?)
You COULD suddenly see the answer in a moment of inspiration - there's a similar problem involving flies on trains, or a man walking home with his dog and various other scenarios that you might have heard of (if none of those sound familiar, don't worry.)
OR you could admit that you don't know what you're doing and start to play around - you are trying to calculate the number of bounces (let's call that n) based on the speed (v), depth (h), width (w) and since this question looks like projectile motion we'll add in gravity (g).
What's you first guess at an answer? Well, it should be something like: n = (hv)/(wg), that is, it crosses more times in a tall well if it's moving fast, than it does in a wide well with strong gravity. Unfortunately, that answer is not only wrong, but it doesn't make any sense: the two sides have different dimensions (units). The way to fix this is to simply allow the numbers to be raised to some powers (see photograph). You need the right hand side to have no reference to time in it, or distance. Dealing with time first, the numerator has v, a speed, and the denominator has g, an acceleration. The quantities v^2/g and v/sqrt(g) both have the time "cancelling out" but which to pick? The answer is that it doesn't matter, but that it's easier to pick sqrt(g) in this particular problem.
Query: Why did we choose the square root when normally integers are easier to work with? Well, it's because in the back of our head we're remembering our failed attempt to work out the distance travelled and there we used s=ut + 1/2 gt^2. When working out times we're going to get factors of sqrt(g). It almost certainly won't matter though - v^2 after all is connected to kinetic energy and g to weight and hence to work done. Just choose quantities that you recognise!
Filling in the same thing for distances we get the following suggestion:
n = h^(1/2)v / (w g^(1/2))
Again, we now do something to make it look more familiar: we regroup that to v/w * sqrt(h/g). That's certainly reasonable. Now, at this point we have a formula that looks intuitively "about right" and that's dimensionally correct. However, we could square the whole thing and get a different formula that would still satisfy those properties. We now need to check - this is where the beauty of the whole scheme lies:
What are the dimensions of v/w ? Well, it's a speed divided by a distance so that's a frequency. (You might think of v=wr from circular motion, or simply cancel dimensions in your head). What about sqrt(h/g)? Well we know it has to be a time so that n is dimensionless, but let's check anyway:
[h]=L , [g] = LT^-2 , [h/g]=T^2, [sqrt(h/g)]= T
So we have that the number of bounces = a frequency * a time. A lightbulb goes off and we suddenly know how to solve the problem PROPERLY. We do the following steps:
1) Find how often the ball crosses the well. We immediately notice that our suggested answer of v/w is actually correct
2) Find how long the ball takes to fall. We use h=1/2gt^2 and get sqrt(2h/g)
3)Combine them, to get n = sqrt(2) (v/w) sqrt(h/g)
So we were a factor of root 2 off. If we had picked v^2 then we would have had more interesting options available - one choice is simply to realise that you have a "frequency-squared" time a "time-squared" and to square root immediately. It's also tempting to try and use an energy argument at this point, multipling in by 2m/2m to get n = 2h/w^2 * (mv^2/2 \div mg). As far as I can see that still doesn't work, but it might be possible to do something clever (nothing springs to mind. At any rate, that formula currently gives n^2, so it's difficult to see why it might help, but maybe you'd strike on the idea of getting time out of power and gravitational potential or something similar that gives you the first approach agian.)
The power of dimensional analysis is fickle: some problems are just impossible to solve this way. Almost anything given in a mechanics test (e.g. M2) won't be doable like this because they've put so many fiddly complications in. But if you have an ostensibly simple situation and are stuck the following steps might help:
1) List all numbers that are relevant. That's all nubers given in the question, g if it's a projectile question, epsilon if it's an electric question etc.
2) Write out a "guess" formula that includes whether things increase or decrease your variable. E.g. maybe you're trying to find the force on a dipole - this should increase with dipole moment and field strength. Since dipoles experience no force in a uniform field, the force law should decrease with the characteristic length scale of the field - it it changes rapidly over short distances the force will be large.
3) Find exponents that get the right dimensions
4) Rewrite in a way that looks prettier
5) Look for inspiration
6) Repeat
