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Title : Solutions To Special Relativity Problems (5)
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Solutions To Special Relativity Problems (5)

1)   A student proposes to compute the kinetic energy of a particle relativistically by using the expression  1/2  m v²  with the 'relativistic mass of the particle.   Would this be correct? Explain why or why not.

Solution:    The student's proposal  must be incorrect given the relativistic formulation for kinetic energy is:

E _K  =    m_o  c²/ [(1 - u²/c²)^½ ] -   m_o c²


I.e. bearing a rest energy (m _o c² ) which is non-zero.  On the other hand, for the non-relativistic case:

E ' _K  =    1/2  ( m v² )   -   1/2  ( m v _o  ² )

The rest energy (2nd) term is zero because the velocity is 0.  This also means the total energy is greater for the relativistic case, i.e.

E _K   +   m_o c²    >    E ' _K    =    1/2  m v²


2) Determine the energy required to accelerate a proton from 0.25c to 0.50c. 

Solution:  By the work -energy theorem:

W = K(f) - K(i)

K(i) = m_o c²/ [(1 - v ²/c²)^½]

v1 = 0.25 c

K(f) = m_o c²/ [(1 - v ²/c²)^½]

v 2 = 0.50c   (where: m_o = 1.7  x 10 ^-27 kg )

Then:

K(f) - K(i) = m_o c²/ [(1 - (0.50c)²/c²)^½]    - m_o c²/ [(1 - (0.25c)²/c²)^½]

K(f) - K(i) = m_o c²/[(1 - 0.25]^½ - m_oc²/[(1 - 0.0625]^½

K(f) - K(i) =  1.15 m_o c²-    1. 03 m_o c²     =  0.12 m_o c²    



But the proton rest mass energy in MeV is:   m_o  c²  =  938 MeV

So that:  0.12 m_o c²      =  (0.12) 938 MeV  =   112 MeV


3) Protons emerge from a particle accelerator with a kinetic energy equal to 0.49 mc².   What is the speed of these particles? Compare the result to that obtained from the non-relativistic relation between mass and energy.

Solution:  Because the kinetic energy is relativistic, the velocity must be as well, so we use the relativistic form for KE:

E _K  =    m  c²/ [(1 - u²/c²)^½ ] -   m  c²

Whence:   

 0.49 m  c²   =   m  c²    /  [(1 - v ²/c²)^½ ] -   m  c²


And:  1 .49 m  c²  =   m  c²/ [(1 - v ²/c²)^½ ]


(1 - v ²/c²)^½ =    m  c²/ 1 .49 m  c²         or:


(1  - v²/c²)   =     (1 / 1 .49) ²  

 And finally:  :   v     =  c Ö {1   -  (1 / 1 .49) ² }   

v = 2.2 x  10 ⁸ m/s

For non-relativistic equation, we get an erroneous   v' =   Ö {2E _K  /  m)  =   2.8 x  10 ⁸ m/s

4)What is the speed of a particle whose kinetic energy is equal to its rest energy? What percentage error is made if the non-relativistic kinetic energy expression is used?

Solution:  Here we have in the KE equation:   E _K  =    m  c²


So that:    m  c²   =    m  c²/ [(1 - u²/c²)^½ ] -   m  c²


Hence:    2 m  c²   =    m  c²/ (1 - u²/c²)^½  

(1 - u²/c²)^½    =    m  c² /  2 m  c ²      =  1/2


u²/c²     =    ( 1   -  1/2 ) ²      =  (0.5) ²   


     u =  c Ö {1   -  (0.5)² }=  c Ö (0.75) =  2.6 x  10 ⁸ m/s


For non-relativistic case, we have, by the work-energy theorem:

  K(f) - K(i)  = Wnet   =  1/2  m u' ²   -  0 

Then:   u' =   Ö (2 Wnet  / m) =  Ö {2E _K  /   m) 

u ' =   Ö {E _K  /   m)  =  Ö {2  m  c²  /  m }     =  4.3 x  10 ⁸ m/s

Percentage error =   [(u' -  u)/ u ]  x 100%  = 63%


5) Show that the relativistic  kinetic energy equation: 

E_k   =   m_o c² / [(1 - v ²/c²)^½ ]    -    m_o c²


reduces approximately to 1/2  m v²   when  v << c.

Solution:

Apply binomial theorem to the relativistic factor, e.g.

(1 - v ²/c²) ^-1/2    =    _{1  +}    v ²/2 c²   +   (3 v ⁴/ 8c⁴  ) +    ....

Then, if we require v << c (non-relativistic case) all the terms containing v/c  or higher powers of this ratio can be neglected. Then we are left with (approximately) :

E_k   =  1/2  m_o v ² 

Thus Article Solutions To Special Relativity Problems (5)

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