### Video Transcript

In this video, we’re gonna look at a problem set in the exciting, but as far as I know, imaginary sport of model snail racing.

But first, let’s learn about model snail racing. Well, it’s pretty straightforward really. We have a set of 25 identical-looking wooden models of snails and a slope. Now because it’s pretty difficult to draw them identical-looking, I’ve put most of mine in a big box. Now we pick five snails at random. Pop them at the top of the slope up here. Let them slide all the way down. And the first one to reach the bottom is the winner!

Now the snails don’t have any moving parts. So their progress down the slope depends entirely upon the differing levels of friction between their bases and the slope. They’ve been designed so that each one looks the same. But they have a slightly, more or less, smooth base and will encounter a different amount of frictional resistance from the slope. The effect of air resistance can be ignored at the sort of speeds that they reach. In fact, some people have tried removing the snails’ shells to make them more streamlined. But they reported that, if anything, it just made the snails more sluggish.

Now the problem that we’ve got for you doesn’t require any high-powered maths, just a bit of logic. But before we describe the actual problem to be solved, let’s take a quick look at the maths behind the racing of model snails. We can model the snail as a particle on a rough slope. And that just means that there is some friction between the snail sliding down the slope. Let’s call the angle between the slope and the horizontal 𝜃. The weight of the snail acts vertically downwards with a force in newtons equal to its mass in kilograms times the gravitational constant, 𝑔, in meters per second per second. And we usually use a value of 9.8 meters per second squared for the acceleration due to gravity at the earth’s surface.

Newton’s third law tells us that for every action, that there’s an equal and opposite reaction. So as the snail presses down on the slope, there will be a normal reaction, let’s call it 𝑅, at an angle which is perpendicular to the slope acting from the slope onto the snail. Now we’ve got a right-angled triangle here. Which means that this angle is 90 minus 𝜃. And this angle is also 𝜃. And remember that normal reaction force is perpendicular to the slope. Now because the snail isn’t leaping off of the slope and it’s not crashing down through the slope, it means that this force 𝑅 must be exactly balancing the component of the weight force which is acting in this direction. And that will be 𝑚 times 𝑔 times the cosine of the angle between that force and this direction. So the reaction force 𝑅 is equal to 𝑚𝑔 times cos of 𝜃.

Then because it’s a rough slope, there will be a friction force resisting the movement of the snail down the slope. Now remember, the thing that makes snail racing so exciting is the fact that the snails are actually moving. So that frictional force will have reached its maximum theoretical level. Let’s call it 𝐹 max. And at the scales we’re talking about, it’s going to be constant throughout these snails’ entire journey down the ramp. Now the value of 𝐹 max is going to be 𝜇, the coefficient of friction between the snail and the slope, times 𝑅, the size or the magnitude of the normal reaction force acting on the snail.

So 𝜇 is the slippery factor of the snail. And we’d better hope, for a nice, exciting race that each snail has a different 𝜇. So in order for this snail to slide down the slope, the component of the weight force acting down the slope must be bigger than this 𝐹 max force here. And when that happens, we get a constant acceleration of, let’s call it 𝑎, down the slope. And Newton’s second law tells us that the resultant force acting on the snail down the slope is equal to the mass of the snail times its acceleration. So we can see that 𝑚𝑔 cosine of 90 minus 𝜃 minus 𝐹 max is equal to the mass of the snail times its acceleration. But cosine of 90 minus 𝜃 is the same as sine of 𝜃. And 𝐹 max is equal to 𝜇 times 𝑅. And 𝑅 is equal to 𝑚𝑔 cos 𝜃. Now we can cancel out the 𝑚s on either side of the equation. And we can see that the constant acceleration of the snail down the slope is this expression here, which is independent of the mass of the snail, which is mildly interesting.

Now we can use one of Newton’s equations of motion, where 𝑠 is the distance travelled by the snail down the slope in meters, 𝑢 is the initial velocity of the snail in meters per second, which of course is zero meters per second because we just plonked the snail down at the top of the slope. So zero times the time it’s taken is gonna make this term here equal to zero. And we can use the expression we’ve just worked out for the acceleration.

And then, since all of the snails are travelling on the same slope, the same distance, at the same angle, with the same gravity, the only thing that varies, is this coefficient of friction for each of the snails. Given that our physical slope looks something like this, we can see from the formula there that the snail with the smallest 𝜇 will slide down the ramp the fastest. Which reminds me of possibly the worst math joke of all time. Which cat slides down the slope the fastest? The one with the smallest 𝜇. And on that note, I think we’d better move on to the main business of the day, the model snail racing problem.

Without a timer, and with races consisting of no more than five snails at a time, just noting their finishing positions in the races, what is the minimum number of races needed to find the first, second, and third fastest snails overall?

Now we don’t know the value of 𝜇 for any of the snails. We don’t know the angle of the slope. And we don’t know the length of the slope. So I don’t want you to use any of the maths I’ve just been talking about in order to answer this problem, just a bit of logic. So pause the video now. Have a go at the problem. And then watch through our solution.

Okay, so we could label all of our snails A, B, C, D, E, F, and so on up to W, X, and Y. Then put A to E in the first race and see who finishes first, second, and third. Then we could replace the snails that finished fourth and fifth with F and G. Then run the race again and see who finishes first, second, and third. Then get rid of the snails that finished fourth and fifth and replace them with snails H and I. Then run the race again and replace the slowest two snails with J and K, and so on. Then at the end of the 11th race, we’d know the three fastest snails.

But can you do this in fewer than 11 races? Now it’s tempting to say that we could run five races, then put the winners of each race against each other in a grand final. Now that would certainly help us identify the fastest overall snail. But it doesn’t necessarily tell us about the second and third fastest. For example, if A, B, and C were the fastest three snails overall, then we’d have rejected two of them for the grand final under that system. So six races isn’t the answer. But if we carefully analyze the results after putting them in a table like this, where the rows are the races and the positions are in columns, then maybe we can get out more information than we first thought.

Now, obviously, depending on the speed of the snails, these letters will appear in different orders. But you can see here that in race one, snail A finish first, C second, D third, E fourth, and B fifth. And that tells us that E and B can’t possibly be among the top three snails in the whole competition. Likewise, in race two, snails G and J finished fourth and fifth. They can’t possibly be one of the top three fastest snails. And likewise, any snail that finished fourth or fifth in its race can’t be one of the top three fastest snails. Now row six is the winners race. And we can see that snail A, not only beat everybody in its first race, but it also beat all of the winners from the final race. So snail A must be the fastest snail overall.

Now we can also see from this that snail V finished fourth in that race. So if it’s not one of the top three, then anything slower than V is also not one of the top three. So we can discount V, U, and W from our reckoning. Likewise, with P, it finished last in the winners race. So anything slower than P, certainly can’t be one of the top three. What about snail H? It finished third in the winners race. So, at best, it could be the third fastest snail in the whole pack. So anything slower than H can’t be considered for the top three. So that rules out F and I.

Thinking about snail O, at best, it’s the second fastest snail overall. Now the contenders for third fastest snail must be H, which finished just behind it in the winners race, and N, which finished just behind it in the original race. K, who finished two places behind it in the original race, can’t possibly be the third fastest snail. Lastly then, we know that A is the fastest. Now that gives us five options for the second and third place snails.

So now we can have one final race, the seventh race. And the winner of that race is the second fastest overall. And the second in that race would be the third fastest snail overall. So six races wasn’t quite enough information to tell us about the overall first, second, and third snails. But with that one extra race and a bit of careful analysis, we can identify the top three snails overall. Now, if you tried this on a pack of 25 racing snails yourself, the order of the letters that you get would very probably be different to the order of letters that we got. But it’s the process of running five initial races, followed by a sixth winners race, then the logic and analysis we used followed by the final seventh race, that leads us to our one, two, and three fastest snails in the pack.