^{2 }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}c

^{2}/ [(1 - u

^{2}/c

^{2})

^{½}] - m

_{o}c

^{2}

^{}

^{}

I.e. bearing a rest energy (m

_{o}c

^{2}) which is non-zero. On the other hand, for the non-relativistic case:

E '

_{K}= 1/2 ( m v

^{2}) - 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}c

^{2 }> E '

_{K }=

_{ }1/2 m v

^{2}

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 }c

^{2}/ [(1 - v

^{2}/c

^{2})

^{½}]

v1 = 0.25 c

K(f) = m

_{o }c

^{2}/ [(1 - v

^{2}/c

^{2})

^{½}]

v 2 = 0.50c (where: m

_{o}= 1.7 x 10

^{-27}kg )

Then:

K(f) - K(i) = m

_{o}c

^{2}/ [(1 - (0.50c)

^{2}/c

^{2})

^{½}] - m

_{o}c

^{2}/ [(1 - (0.25c)

^{2}/c

^{2})

^{½}]

K(f) - K(i) = m

_{o}c

^{2}/[(1 - 0.25]

^{½}- m

_{o}c

^{2}/[(1 - 0.0625]

^{½}

K(f) - K(i) = 1.15 m

_{o}c

^{2}- 1. 03 m

_{o}c

^{2}= 0.12 m

_{o}c

^{2}

But the proton

*rest mass energy*in MeV is: m

_{o}c

^{2 }= 938 MeV

So that: 0.12 m

_{o}c

^{2}= (0.12) 938 MeV = 112 MeV

3) Protons emerge from a particle accelerator with a kinetic energy equal to 0.49 mc

^{2}. 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 c

^{2}/ [(1 - u

^{2}/c

^{2})

^{½}] - m c

^{2}

Whence:

0.49 m c

^{2}= m c

^{2}/ [(1 - v

^{2}/c

^{2})

^{½}] - m c

^{2}

And: 1 .49 m c

^{2}= m c

^{2}/ [(1 - v

^{2}/c

^{2})

^{½}]

(1 - v

^{2}/c

^{2})

^{½}= m c

^{2}/ 1 .49 m c

^{2}or:

(1 - v

^{2}/c

^{2}) = (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 c

^{2}

So that: m c

^{2 }= m c

^{2}/ [(1 - u

^{2}/c

^{2})

^{½}] - m c

^{2}

^{}

^{}

Hence: 2 m c

^{2 }= m c

^{2}/ (1 - u

^{2}/c

^{2})

^{½}

(1 - u

^{2}/c

^{2})

^{½}= m c

^{2 }/ 2 m c

^{2 }= 1/2

u

^{2}/c

^{2 }= ( 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 c

^{2}/ 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}c

^{2}

**/**[(1 - v

^{2}/c

^{2})

^{½}] - m

_{o}c

^{2}

reduces approximately to 1/2 m v

^{2}when v << c.

Solution:

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

(1 - v

^{2}/c

^{2})

^{-1/2 }=

_{ 1 + }v

^{2}/2 c

^{2 }+ (3 v

^{4}/ 8c

^{4 }) + ....

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}

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