Energy Transformations (2) EPE to KE: The Battle of Agincourt

A bow stores elastic potential energy in the flexible ash of the wood of the bow and the stretched string.

The English archers won the Battle of Agincourt in 1415 because the range of the better-made and longer English bows was greater than the bows used by their French enemies. They stored and released energy more efficiently.

They could probably count on a release speed of 100 feet per second = 3000 cm/s = 30m/s and they knew that a 45 degree angle gave maximum range.

 

So 30 cos 45⁰ = horizontal speed = 21.2m/s. This stays the same because  a=0 in the horizontal plane.

In the vertical plane, we have to use the equations of motion (constant a). With 30sin45⁰ as u, and acceleration = g (9.81ms^-2)  this  yields t = 2.16s going up, PLUS the same coming down so the arrow’s total time in the air = 4.32s.

So, range =21.2×4.32 = 91.6m…and vertical height reached is  33.2m as long as air resistance is neglected.

Now, let’s take this apart…

At the top of its flight, what kind of energy does the arrow have? Estimate arrow mass at 0.1kg (is this reasonable?)

A:  Using kinetic energy in the horizontal direction (22.47J) plus GPE at maximum height (32.57)

Estimated energy 55J

Where does this energy come from? Think about the area under a F/x graph for a spring or similar where Hooke’s Law is obeyed…

A: the energy stored in the stretched bowstring. Let’s assume it obeys Hooke’s Law.

Using our calculations so far, we can now estimate the maximum pulling force the archer would have to use. Suppose the bow is drawn a distance of 0.5m

With quite a lot of ifs and buts – you can see where the approximations are – the archer would have to pull with a maximum force of 55/0.5N or 110N. This would be like holding a mass of 11kg just off the ground. With one hand…

 

Heisenberg’s Uncertainty Principle. Confining a wave/particle in a box.

In subatomic terms, because of wave particle duality, certain pairs of measurements such as where a particle is (x) and  where it is going (its position and momentum) cannot be precisely known. If we know one very precisely, the other cannot be known. Putting this another way, a particle has mass (hence momentum) also a wavelength given by the de Broglie expression

When particles’ wavelengths interfere, they form a wave packet of finite size having a length which has to fit into the confining box, which can happen in a variety of ways…Here, we’re only really concerned with the smallest “wavefunction”, shown in red at the bottom. The diameter of the box is approximately half a wavelength. The rest are there just to show what’s possible. A wavefunction represents the probability of finding the wave in a particular space.

Confining our wave in a box, where it is and its momentum are defined like this:

This makes sense in the context of a problem. Imagine an alpha particle confined within a nucleus of gold. Given the alpha particle has a wavelength confined by a ‘box’ the size of the nucleus, whose diameter might be:

Suppose we want to find the energy of the confined alpha particle. We use:

The energy can be found using a different expression:

We can find the mass of an alpha particle (2 protons and 2 neutrons. If we plug in the numbers, we get 4.3×10^-15 J or 27keV, consistent with observed energies.

You might try the same calculation to find out the energy an electron would have to have to confine it inside the nucleus.

This is why we don’t get electrons inside nuclei…