Wednesday, July 16, 2008

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.


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

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