QUIZ 26(TEST 4)  AND (TEST 3)

CQ's " Conceptual Problems: 1, 6, 9, 10, 11, 12, 16, 18, 19, 20
Problems:  time dilation; length contraction (It's just a new way of thinkin') -1, 2, 3, 4, 5; "Modern" Momentum-15, 16,  17, 18;  E = mc2- 20, 24, 25, 26, 27, 28; Adding velocities: 43, 44, 45, 46
Turn in:  Problems 2, 4, 16, 20, 26, 45, 46
1. Commit these magic words to memory: Reference- page 740 in BOLD: The length of an object is measured to be shorter when it is moving relative to the observer than when it is at rest. 

Suppose  you are moving in a bus past a bus stop. The stop  will look shorter to you than to a person sitting at rest on the street; likewise,  a card table  on a bus moving  past you while you 're sitting at the stop will look shorter to you than to a person playing Black Jack on the bus. 

In this case, a space ship passes by you while you are sitting at the "bus stop,"  a  metaphor for the reference frame "at rest."  Use equation 26-3. 
2. Commit these magic words to rest: "The time interval for an event (such as two consecutive  heart beats) will appear to be longer when the reference frame "carrying" the event is  moving relative to the observer than when it is at rest. " This is a rather wordy para-phrase of the bold faced sentence  page 735 stating time dilation : "clocks moving relative to a observer are measured by that observer to run more slowly (as compared to the clocks at rest.)"

In this case, use equation 26-1a. Be careful: here we have (delta t) = 4.76 x10 -6 seconds = (delta to)/(1 - v2/c2)1/2  . You are solving for delta to. 
16. mo*c2/(1 - v2/c2)1/2   = 2*mo*c2 .   Here I equated the total energy to  twice the rest energy. This is the same  as equating the relativistic mass to twice the rest mass as required:

mo/(1 - v2/c2)1/2   = 2*mo.   In any event, manipulate this to get:

(1 - v2/c2)1/2   =  1/2.  Square both sides,
1 - v2/c2   =  1/4, 
and solve for v.  
20. E = mo*c2/(1 - v2/c2)1/2  .  Assume v = 0, since the input chemicals  and reaction  products are at rest. 
Initially, before the reaction Eio = mio*c2 . After the reaction: Efo = mf0*c2 , where mio and mfo are the initial and final rest masses, respectively.  Efo - Eio = (mf o- mio )*c2   . Find the  mass change from the  energy difference. 
26.  KE = difference between total energy when moving and  to the total energy when at rest =  Etotal - Eo = 
mo*c2/(1 - v2/c2)1/2   = 2*mo*c2 .   The mathematical manipulations mimic problem 16's. 
Solve for v this way:
1/(1 - v2/c2)1/2   = 2 .  Guess what. This expression is problem 16's.  For v, invert and square both sides of the equation. 
45. 
(a) See my class notes; this part mimics exactly example 26-16. 
(b) This is a little thought provoking but take a look at figure 26-10. Assume rocket 2 is fired in the opposite direction,  toward Earth. If our Galilean, pre-Einsteinian intuition prevailed, we'd conclude that u = 0.71c - 0.87c. But alas, we'd be wrong since we must insert a "corrective" denominator  preventing relativistic violations.

Remember, c is the ultimate speed. Without the denominator, you might compute speeds > c. Or you might compute v = c, which  is theoretically impossible for particles with non-zero rest mass.  So we must write:
u = (0.71c - 0.87c)/[1 + (0.71c)*(-0.87c)/c]. Here the corrective denominator = [1 + (0.71c)*(-0.87c)/c]. Note  the positive direction of motion is considered to be away from Earth. Thus, the velocity of rocket 2 is negative (-0.87c) since it's launched toward Earth. 
46. This problem is  little thought provoking but take a look at figure 26-10 and fig. 26-12 to compare. In figure 26-12,  rocket 1 is moving  toward Earth. Rocket 2 is also moving toward Earth. Both speeds relative to Earth are given, 0.60c and 0.90c, respectively.  In this case our Galilean, pre-Einstein,  intuition concludes that rocket 2 relative to 
rocket 1 has velocity  = 0.90c - 0.60c. Think of it this way. You are moving at 0.60c relative to Earth and a car behind you is you is traveling at  0.90c relative to Earth. Classically,  you would say the car behind you is moving toward you at  
u = 0.90c - 0.60c.  

Folks do this everyday on Highway 880.   For example, if 2  cars are moving at the same velocity relative to the exit, the occupants  of one car perceive the other car as at rest.   

But alas, we be in error to say u =  0.90c - 0.60c because of the "corrective" denominator. 

So we must write:
u = (0.90c - 0.60c)/[1 + (-0.60c)*(0.90c)/c]. Here the corrective denominator = [1 + (-0.60c)*(0.90c)/c].  Note the positive direction is now toward Earth. Clearly, if I was on the Enterprise (rocket 2) I would see rocket 1 traveling toward me, away from Earth; hence the negative sign in front   of 0.60c.

Another perspective:  See equation 3-6, Chapter 3, page 63.  Note the subscript placements. In 1-D,  we wrote in Physics 2A: V21 = V2E + VE1, where V21 is the velocity of rocket 2 relative to rocket 1, V2E is the velocity of rocket 2 relative to Earth, and  VE1 is the velocity of Earth relative to rocket 1.  This is a mindful way to formulate the problem of relative velocities,  keeping track of subscripts this way. Note that "E" in the last equation "disappears."  

Clearly, V2E is positive. V2E = 0.90c  .  But VE1  is negative. It must be since to an observer on  1, the Earth seems to be moving toward  1,  in the negative direction.VE1  = -0.60c.