Part 2: Explaining the data with special relativity Recall that we derived the formula K = (y-1)mc in lecture by applying the work-energy theorem. You will now show that this formula can explain the experimental data from Part 1, and that it reduces to the Newtonian result in the appropriate limit. You will need the Taylor series (1+ x)= 1+or + "(@ -12 +0(23). 2 2.1. Rearrange the relativistic kinetic energy formula to find v2 as a function of K. Plug in a few values for K between 0 and 2.5 MeV and draw the corresponding values of v on the experimental data plot (reproduced again below): how well do they agree? v2 (1015 m?/sec?) ONADOC O 0.5 1.5 2 2.5 Kinetic energy K (MeV) 2.2. To see the limiting speed, take the limit of your formula as K - co. Do you recover your answer from problem 1.3? 2.3. To recover Newtonian mechanics, take the opposite limit: Taylor-expand (y - 1)mc in the small dimensionless parameter v2/c', and check that you recover your result from problem 1.1 to leading order in v2/c2. 2.4. As we take more and more terms in the Taylor series, we can approach the full special relativistic result. Show this by plotting the first relativistic correction to the Newtonian formula on the plot from problem 2.1. Your answer should take the form K = =mev + Cul, so your job is to find the constant C. At approximately what speed v does the result begin to differ significantly from the data? 2.5. Since K is small whenever v is small, we could just as well have performed a Taylor expansion in K. However, K is not dimensionless, so we have to compare it to something else with units of energy. A good choice is the electron rest energy En = m.c'. Show that a Taylor expansion of your formula from problem 2.1 in the small parameter K/Eo gives the correct Newtonian result v' = 2K/me. The implication of this calculation is that Newtonian mechanics is a reasonable approximation whenever the kinetic energy of a particle is small compared to its rest energy.Part 3: A first look at relativistic kinematics with photons We have already seen in several ways that it takes an infinite amount of energy to accelerate a massive particle to the speed of light. A massless particle would make no sense in Newtonian mechanics (since all of the kinematic formulas like F = ma and K = =mv' would just give zero), but such particles are important features of special relativity. The most common example is the photon, which is a particle of light. 3.1. Take the ratio of the relativistic momentum and energy formulas to derive a useful formula for the velocity of a particle: p/E = v/c2. Notice that this formula does not involve the mass! 3.2. Suppose we boldly assert that p/E = v/c still applies when v = c. In that case, we could still assign an energy and a momentum to a photon, related by Pphoton = Ephoton/C. Rearrange this formula to show Ent Ephoton - Pphoton = = 0. 3.3. Use the relativistic momentum and energy formulas for a massive particle to show that E2 - pac = me. This is probably the single most important formula for the remainder of the course! And by your work in problem 3.2, it also applies for a massless particle with m = 0. 3.4. Since a particle at rest has E = Eo, it can convert its rest energy into other forms of energy. In particular, there is an unstable subatomic particle called the 7 (pronounced "pi-zero" or "neutral pion" ) which decays into two photons. Use the results of problems 3.2 and 3.3 to determine the energies of the two photons in terms of the mass of the pion my. To do this, you can work in the rest frame of the pion, and assume that the photons' momentum vectors are aligned with their velocity vectors (this will be fully justified in a few weeks). First apply momentum conservation, followed by energy conservation