Discuss the below:
1. Suppose we accelerate a proton so that its kinetic energy is equal to its mass energy. What is the speed of such a proton? NOTE: you do not have to know the mass of the proton to do this problem: it would be the same for any particle.
2. Spaceman Spiff passes the earth on his way to Mars. His thruster control is stuck and his spacecraft is traveling at a speed 0.95c. (That is, 95% the speed of light.) The distance from the earth to Mars at that time is 1.2 x 10^11 m.
(A) How long (as observed by someone on the earth) will it take Spiff to travel from the earth to Mars? (B) How much time will Spiff measure for his trip from the earth to Mars? (In his own frame, Spiff is at rest, so this may be considered the time for Mars to reach Spiff.) (C) What is the distance Spiff measures for the distance from the earth to Mars? (D) Suppose Spiff's spacecraft has some mass M. In terms of the rest mass energy, Mc^2, how much work had to be done on Spiff's spacecraft to accelerate it to 0.95c? (By the work-energy principle, this is equal to the craft's kinetic energy.) (E) Suppose Spiff has a very light spacecraft, with a mass of only 1000 kg. What is the work done to accelerate it up to speed, expressed in joules? Compare this to the yearly energy consumption of the United States, about 10^20 J.
3. Two clocks, "A" and "B", are synchronized in their own reference frame. A third clock, "C", is traveling very fast to the left, as shown in the diagram. When "C" passes "B", the clocks display exactly the same time reading.
(a) When "C" passes "A", which clock will be ahead of the other?
(b) As we stated above, clocks "A" and "B" are synchronized in their own reference frame. Are they synchronized according to clock "C"? That is, in clock "C"'s frame, clocks "A" and "B" are moving to the right. Does "C" observe the two clocks to have the same reading? Explain clearly. (Hint: just changing frames cannot change "events". When "C" and "B" pass, they will show the same time reading no matterwhose frame you are in. Similarly, if "A" is ahead of "C" in one frame when they pass, it will also be ahead of "C" in any other frame when the two clocks pass.)