Senior Physics Challenge Day 5

Today was the last day of the Senior Physics Challenge, and now I'm writing this not at Newnham House, in Cambridge, but at home at my desk. Still, I should finish up what I started, and that means telling you what I did today.

Classical mechanics focussed on momentum, with two main questions of interest today- the first was phrased in true Cantabrigian fashion: two students stand on either side of a 5m punt and one throws a sack of potatoes to the other. How far does the punt move whilst the sack is in flight?

The momentum based answer is to use the mass of the potatoes (20kg), the speed of the potatoes(2m/s) and the mass of the punt (980kg) to solve for the motion explicitly. By conservation of momentum the punt must travel at 2*20/980 = 2/49 m/s. The potatoes have a relative velocity of (2+2/49)m/s and so get to the other end of the punt in 2.45sec, where the other student catches the potatoes, transferring the momentum to the punt and bringing it to rest. In those 2.45seconds, the punt has been travelling at 2/49m/s and so has travelled 0.1m.

Another option is to use a classic result of Newton's 3rd law, which is that external forces cancel out, and so the centre of mass of an object can only be accelerated by external forces. There are no external forces in our problem, so the centre of mass does not move relative to a stationary observer. Using this, we can find the centre of mass before the sack is thrown to be 0.05m from the centre of the punt and by symmetry is must be 0.05 m on the other side of the centre afterwards. Since the CoM has to remain still relative to a stationary oberver, the centre of the punt must have moved 0.1m, exactly the same as before.

The next interesting question was about collisional drag on satellites. The basic premise was to find the drag on a satellite. We can do that as follows: the velocities of air particles is pretty much random and is zero on average, so they will not push on a stationary satellite. However, the moving satellite causes the air particles in front to have a positive velocity relative to the satellite, giving them net momentum. If we assume that the satellite moves through a volume V of air, then we argue the following: the particles will collide with the satellite and they will lose on average p=mv momentum. Sometimes they will lose 2p=2mv, because they rebound totally elastically in the opposite direction, sometimes they won't lose any. The average velocity of the particles is just the velocity of the satellite remember, so we can say that:

dP = dmv = dhVv, where h is density.

dP = dLAhv, where L is the length of the volume that the sat has passed through and A is the sats cross section

dP/dt = dL/dt (Ahv) = vAhv = Ahv^2 = F

For the specific case of a satellite, we can do better than prove the dependance oon v^2 - we can find the value and plug it in. The acceration will be given as:

v^2/(r) = g = GM/(r)^2

v^2 = GM/(r) , where r is height from the Earth's core, so r= altitude + Earth radius

therefore, F = GMAh/(altitude+Earth radius)

In Quantum mechanics we looked at finite wells - in particular finite square wells. These are more complicated that infinite square wells because of a finite probability of finding the particle outside the potential well, in a region where it would be classically impossible to reach (i.e. it has to have negative kinetic energy to be there). This also screws with your nice boundary conditions where you demand that w(x) vanishes at the boundaries, because that's no longer the case.

In particular, the schroeinger equation for this case is given as:

-hbar^2 D^2 w(x) + V(x)w(x)= Ew(x)

Inside the well that has a solution D^2w(x) = -k^2w(x) --> w(x) = Asin(kx + C)

Outside the well, you first need to bring the V to the other side. Calling the height of the square well V, that gives you:

-hbar^2 D^2 w(c) = -(V-E)w(x)

For the case V>E, which is when the well is actually a well, (I.e. the particle has less energy that the size of the potential drop at the ends), the energy eigenvalue -(V-E) is negative. We get a solution:

D^2 w(x) = k'^2 w(x) --> w(x) = Be^(k'x) and w(x) = Ce^(-k'x)on either side of the well. It seems reasonable that B=C due to symmetry when the solution inside is an even function like cos, and B=-C when the function inside is an odd function like sin.

The boundary conditions for this are that w(x) must be continuous at the boundary and that the derivative must be continuous at the boundary. That gives you some rather ugly equations which have to be solved numerically, but now at least you have the solution for a square well of any size at all! Now just to solve for all the other possible well shapes...