QUIZ 2 CH. 37---
discussion questions--9, 10, 13
     exercises--37.5: 14, 16, 22, 23; 37.7: 27, 29, 30; 37.8: 43, 44.
     problems---Note: They  tend to combine (i.e., synthesize)
the above 3 sections-49(b) only, 55, 56, 58, 59, 60.
EVEN ANSWERS
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14. Involves  is a simple algebraic manipulation of the following linear equation: 
x' = ( x -  ut)/(1 - v2/c2)(1/2)   
t'  =  (t -  u*x/c2)/(1 - v2/c2)(1/2)   

REWRITE AS 
x' = gamma*(x -  ut)   = gamma*x - gamma*ut
t'  =  gamma*(t -  u*x/c2) = gamma*t  - gamma* u*x/c2  , 
where  gamma = 1//(1 - v2/c2)(1/2)   

Note, the linear system,
   
(I)  x' = gamma*x - gamma*ut
(II) t'  = gamma*t  - gamma* u*x/c2  

can easily be "inverted" to find x and t using MATRIX OR algebraic methods. Note: The Lorentz Transformation is a linear transformation  like some of you already covered in Math 6. A  4-dimenional space -time coordinate system 
 x, y, z, t  is "rotated"  and/or "translated" relative to the the 4D system x', y', z', t'.  A 4x1 column matrix  representing
x, y, z, t  is transformed into the 4x1 matrix for x', y', z', t' via a 4x4 matrix which you can easily construct from  basic linear algebra. (You don't even need Math 6 to actualize  this process depending on algebra/ trig exposure.) 

To do this the old fashioned way , let's say you wanted to get x in terms of x' and t'.  You would multiply equation (II) above by u and add the two equations (I) and (II) to eliminate  gamma*ut and immediately find x in terms of x' and t'. 
This technique can be directly applied to find t in terms of x' and t'  but you'd multiply equation (I) by
u/c2 then add equations, eliminating x.  
FOR 16, 22, and 23, see lecture notes, check back later. 
27.  (a) Referring to components along the  x axis. ax = dvx/dt.  where Fx = dPx/dt  = d[mvx/(1 - vx2/c2)(1/2) ]/dt . See page 1290. Start like this:  d[mvx/(1 - vx2/c2)(1/2) ]/dt  =  [m/(1 - vx2/c2)(1/2) ]*dvx/dt +   mvx*d[(1 - vx2/c2)(-1/2) ]/dt.   Differentiate the second term on the right carefully.  Use the chain rule: 
Second term hint: ( -1/2)*mvx*(1 - vx2/c2)(-3/2)   *  (-2vx/c2)*dvx/dt.  

From this hint you can  put the two terms over a common denominator and  then manipulate the numerator so terms cancel,  leaving only the acceleration multiplied by a constant .
(b)  In this case we have a conceptually simpler problem, since v is constant; you essentially  the derive expression of Chapter 3's uniform circular motion.
vector-F = d [vector-P]/dt  = md[vector- v/(1 - v2/c2)(1/2) ]/dt . =  m/(1 - v2/c2)(1/2) ]*d[vector- v]/dt.   For uniform circular  motion  the magnitude | d[vector- v]/dt |  = v2/r,  where r is the radius of motion. 

29.
(a) magnitude P =  mv/(1 - v2/c2)(1/2)   = 2*mv. Solve for v by squaring both sides.
(b) ma*( 1 - v2/c2)(-3/2)  = F     .    When the  particle is at rest : ma = F.  We have: ma*( 1 - v2/c2)(-3/2)  = 2*ma
30. (a)  F = ma*( 1 - v2/c2)(-3/2)    (b)  F = ma*( 1 - v2/c2)(-1/2)  , where v is constant as in uniform circular motion.
43. (a) integral of  Fx*dx = q*(change in potential)  = mc2/ ( 1 - v2/c2)(1/2)   - mc2 , where v = 0.980*c. Find the change in potential, where q = electron charge magnitude and m is electron's  mass.
(b) Multiply part (a)'s result by  charge magnitude e = q.
49(b) Etotal =  mc2/ ( 1 - v2/c2)(1/2)      
55.
(a) 7 TeV = mc2/ ( 1 - v2/c2)(1/2)   - mc2,  where 1 eV = 1.6x10-19 J. and T = Tera = 10 12.
(b) relativistic mass = m/( 1 - v2/c2)(1/2)
56. Mc2 = (1 - 10 -4 )*Mc2 + Energy Released.   Find Energy Released. Note: Power = Energy Released/time and Energy Released = M'gh, where M' is the hypothetical mass to be lifted by the enormous energy this nuclear bomb releases. This problem is a scary testament to the dangers of nuclear weapons proliferation. The deciding element  may be public pressure so please stay informed. Future scientists and engineers can play an important role explaining some of the technical issues but ultimately it's a  political question.  Albert Einstein  and "father of H-bomb" Edward Teller   were on opposite political poles and had radically different views on nuclear warfare.

This problem also creatively contrasts with the  nuclear model we'll encounter later---Ch. 43, see example 43.3 dealing with the "mass defect.".  We'll find  a nucleus is made up of nucleons (protons and neutrons) "bound" by the attractive nuclear potential. Thus Mc2 = sum of individual nuclear mass - |U|, where |U| is the magnitude of the 'binding" energy. In that  case we see the total rest mass of  nucleus is less than the sum of the individual  rest masses  because the negative  nuclear attractive  potential energy holds the system together. The system is like a planet (mass M) orbiting the Sun (Mass Ms): Energy = sum of kinetic energies - GM*Ms/r, where GM*Ms/r  is the "binding" potential energy. (See Chapter 12. }
58.  Note: A photon, an electromagnetic "particle"  which we'' discover in Ch. 38,  has no mass; it thus travels at the speed of light c. Assume the initial momentum of the system is ZERO; in other words the  atom is at rest before the decay:
conservation of momentum says, 

0 = Patom - Pphoton, where we assume the particles, atom and photon,  move in opposite directions. We just subtract momentum magnitudes, hence the minus sign between two positive numbers.
From equation 37.40 and related textbook discussion, Ephoton = Pphoton*c; thus:
 Patom = Pphoton = E/c.

Relativistically, we have Patom =  mv/( 1 - v2/c2)(1/2)    = E/c. You  could easily find  speed v by squaring both sides and solving for v in terms of m and E. However the book wants you to ignore v/c  and find v in one simple step assuming essentially v/c = 0.  I would require you  to also find v relativistically on exams.
59.
(a) This problem will help you understand #22 and 23. If you want, use my notation without extraneous negative signs, then make the velocities positive or negative as needed: Here is the template, borrowed from Physics 4A:
VAC =  (VAB + VBC)/( 1 +  VAB*VBC/c2) .   This reads the velocity of A with respect  to C = 
velocity of A relative to B + velocity of B relative C,  divided by a relativistic factor given by  ( 1 +  VAB*VBC/c2).  
We have two protons moving in  opposite directions. Let A be the reference  frame for one of the protons and C be the other proton's  frame.  B is the lab reference frame. Let the rightward direction be positive and let proton A be moving rightward relative to the lab; thus VAB  = 0.500c > 0  and    VBC  =  0.500c since VCB = - 0.500 c because Proton C moves leftward relative to lab.
(b) and (c): Check back later. 
60. Check back later and/or see #59's discussion. #60 is a follow up to #59, like "part (d)."