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 = mo c2/
[(1 - u2/c2)½ ] - mo 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 + mo 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) = mo c2/ [(1 - v 2/c2)½
]
v1 = 0.25 c
K(f) = mo c2/ [(1 - v 2/c2)½
]
v 2 = 0.50c (where: mo = 1.7
x 10 -27 kg )
Then:
K(f) - K(i) = mo c2/ [(1 - (0.50c)2/c2)½
] - mo c2/ [(1 - (0.25c)2/c2)½
]
K(f) - K(i) = mo c2/[(1 - 0.25]½ - mo
c2/[(1 - 0.0625]½
K(f) - K(i) = 1.15 mo c2
- 1. 03 mo c2 = 0.12 mo c2
But the proton rest mass energy in MeV is: mo c2 = 938 MeV
So that: 0.12 mo 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 {2 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:
Ek = mo c2 / [(1 - v 2/c2)½] - mo 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) :
Ek = 1/2 mo v 2
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