2nd Round

This is the second question of the 1st IPhO - try it for kicks! (Click on it to view the enlarged picture - same for all other pictures in this post!)

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I actually remember doing this in Physics ‘S’ paper! Haha. So let’s give it a try, using my well, noob method. Let’s consider a simple system at the start first:

Obviously the resistance here is just the simple sum of two resistors in series:

Let’s continue to add two more resistors; notice that I’ve renamed one of the resistors x (you’ll know why in a second!):

This is a little trickier; the resistance here must take into account the parallel circuitry, which I’ve circled out in red. If you can’t see this, well, let me redraw it in a more accessible manner:

And of course, if you still remember the formula for parallel resistors, this new set-up has a fairly easy to calculate expression for its resistance:


And then let’s do one more addition of a branch to see how everything adds up, to understand how I’m going to simplify the situation:

Well, without redrawing it, I’d expect you to say that the set-up now consists of two parallel set-ups. For those who can’t see yet, I’ve tried to point out using circles again. Look above at the blue circle; I can represent this as a resistor x, and then the diagram is simplified into:

And what do you notice? Haha, we’ve already obtained the resistance for this, which was determined earlier to be:

So what’s x? Well, let’s find out; referring to the blue circle, I see that x’s resistance is simply equivalent to a set of parallel resistors:

Okay, so that you’ve roughly got my trick of simplifying, let’s do a long set so that you’ll be seeing what I’m seeing:

Let’s do the red circle first, and let’s call this new resistor a:

Now, treating the red circle as a resistor a, we can simplify the diagram into:

Alright, so now let’s do the blue circle and let’s call this new resistor b:


So now we can simplify again:


And now let’s do the green circle and let’s call this resistor c:


And now, let’s simplify again:

This couldn’t be simpler, and let’s do the orange circle this time and call it a resistor d, and find its resistance:

With this, we can then simplify our diagram as:

And the overall resistance is therefore simply:

Now, if you haven’t seen or noticed the pattern yet, let me just write out everything in full:


What a monster! You may want to view it in full size, haha. Notice that there is a repeating unit! And if the series goes on and on indefinitely, let me just re-write it in a more digestible way:


Can you spot the repeating unit yet? Haha. If you haven’t spotted the recurring unit yet, it’s simply this monster right here:

But what is this creature? Notice that the overall resistance R is just the repeating unit:

So how do we solve this thing? Well, algebra gives us a good method:

And then now we manipulate this expression into a quadratic equation, as such, to obtain R:

Which can then be solved using the standard quadratic equation (notice I’ve rejected the negative answer since no resistance is negative):

And as some of you might have already noticed, the Golden Ratio (phi) appears in the answer:

Such that:

Well, what is the Golden Ratio? Let’s leave that for another entry, because I’m absolutely tired of typing out equations and expressions for the day. I’ll give a hint though; recall the Fibonacci sequence as:

If you look closely, each succeeding term is obtained via addition of its two preceding terms. Well, if you take any term and divide it by its preceding term, you’ll obtain a number close to the Golden Ratio (phi). Let’s say we take 89 and 55:


And this ratio gets increasingly closer to the Golden Ratio as one continues on into the series.

And wow! What a lengthy post! And it's rather strange isn't it? Then adding more and more resistors forces your resistance into a fixed, finite value instead of an infinite resistance, and the fact that the Golden Ratio pops up in such an instance! Simply amazing! :p

1st IPhO Ever!

I'm very sure everyone has at least heard of the International Physics Olympiad, an annual event that has its origins in the year 1967, which targets brilliant high school students who have an aptitude as well as an attitude for Physics.

Unfortunately, yours truly wasn't one of those brilliant students. :( Haha.

But in any case, I've gotten my hands on the first IPho paper, which was held in 1967, Warsaw, Poland, and here's the first question:

If you can't see the picture clearly, please click on it for a full-sized image.

But anyway, the first few olympiads weren't too tough, in my opinion - how do I know? Well the fact that I can at least do them shows that they're not as hard as the recent ones! However, the recent ones tend to be more guided.

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Neglecting air resistance, we can consider our system, comprising of the ball and the bullet (of course, with the inclusion of the Earth), to be not acted on by external forces – in such a system, we can then apply the Principle of Conservation of Momentum, which states that:

The total momentum in an isolated system remains constant before and after any event.

With this statement, we can then say that:

Initial Momentum = Final Momentum

What is the initial momentum? Well, at the start the bullet only has a horizontal velocity component and the ball is stationary, so we say that the initial momentum vector is towards the right, and is solely possessed by the ball. Specifically, we are comparing the total momentum along the horizontal axis. The final momentum must then be compared along the same horizontal axis, and thus we now have to consider the horizontal velocities of both the ball and bullet.

So let’s start by calculating the initial momentum of the bullet:


Calculating the final momentum of the system is a little more tricky and involved; first of all, let’s focus only on the horizontal component of the ball’s velocity, and notice that the horizontal velocity is in no way affected by gravity, such that it remains at a fixed magnitude throughout the descent of the ball.

Notice also that the horizontal distance through which the ball falls is determined by the time of descent as well as the horizontal velocity; and we are given the horizontal distance, and thus we need only determine the time of descent to figure out the horizontal velocity. So we proceed:


Where h is the height given in the question, u is the initial vertical speed (which is zero) and g is acceleration due to gravity. And therefore, we have:


With this, the horizontal velocity of the ball is then:


And now, with this piece of information, we can then determine the final momentum of the system:


Where v is the final horizontal velocity component of the bullet. Of course, by equating this to the initial momentum, we can determine v:


Now, we figure out the time of descent of the bullet, using the same equation of motion:


And well, it should’ve occurred to you that the bullet should reach the ground the same time as the ball does actually, since there is no air resistance, and all objects dropped vertically will reach the ground at the same time. So, we say that the horizontal distance, x, travelled by the bullet after collision is given by:


Amazing, the bullet travels 103 metres! A staggering distance as compared to the more massive ball (20 m).

Now, to find the amount of kinetic energy lost as heat, we simply find the difference between the total final kinetic energy and initial kinetic energy:


That simply means that 1158 J of energy was converted from bulk kinetic motion of the bullet into heat transferred to ball, and thus the fraction lost is then:


Gosh, that’s a mighty sum! :p