Comments on Test 2

1. Most of you did not recognize that this was a chance to think of relativity, that is, to invoke the equivalence of all inertial reference frames. In particular, the second collision, observed from the reference frame of the driver moving relative to the ground, looks exactly like the first collision, to that driver.

2. What most of you did not say, and I wish that you had, is that because the muon's lifetime is a proper time interval, you can use the time dilation formula. Remember that you are supposed to explain your reasoning.

3. Most of you did very well on this one. The only difficulty for some came when they used the velocity addition formula, and switched the roles of the muon's velocity and the neutrino's velocity. Maybe they didn't feel right putting a minus sign in front of the c, but that's what they ought to have done. Then, for the southbound neutrino, they would have found a velocity of -c, indicating that this one is travelling south, not north.

4. Ouch. Well, those of you who thought to use intervals had a pretty easy time of it, easily showing that there can't be any reference frame in which the two clocks agree. Many people tried to use Lorentz transformations, and got into a lot of trouble (For example, many people "found" a reference frame in which the two clocks agreed.) by not being clear about what each event was, or at least, what x and t were in each frame.

5. Again, most of you did quite well - after all, this was very similar to a sample problem. The most frequent shortcomings included: not explaining your answers (e.g., why does Zork receive the "I Love Lucy" broadcast in the year 2055?), messing up your algebra, and failing to account for time dilation in the last question. One thing that people did to make our lives harder was to convert speeds into m/s and time into seconds and distances into meters. This problem is a perfect example of its being sometimes best to leave distances in lightyears, time in years, and speeds in multiples (fractions) of c.

6. Most of you correctly noted that a negative interval between B and C, in one frame, meant a "timelike" separation between these events, in which case there would be no reference frame in which their order was different. Those who got into trouble calculated the interval mixing values from two reference frames, which is never OK. Also, quite a few people asserted that C occurs after A in the primed frame (nope). Finally, many of you messed up your arithmetic. I was pleased to see some of you using intervals to calculate when C occurs in the primed frame, for example. Good for you!