Posted by **martini** on Saturday, April 28, 2012 at 2:57am.

A particle has a kinetic energy equal to its rest energy.

(a) What is the speed of this particle?

(b) If the kinetic energy of this particle is doubled, does its speed increase by more than, less than, or exactly a factor of 2?

(c) Explain

(d) Calculate the speed of a particle whose kinetic energy is twice its rest energy.

I got (a) and (d)

(a) .866c and (d) .943c

but I'm stuck on (b) and (c) I think it might be less than 2 but I have no idea why that would be the case, can someone please explain it to me!? Thank you!

- physics -
**Elena**, Saturday, April 28, 2012 at 7:11am
(a)

KE = m(o) •c^2•[1/sqrt(1-β^2) -1],

E(o) = m(o) •c^2,

KE = E(o),

m(o) •c^2•[1/sqrt(1-β^2) -1] =

= m(o) •c^2,

1/sqrt(1-β^2) =2,

β = v/c = 0.866.

v = 0.866•c =0.866•3•10^8 =2.6•10^8 m/s.

(b),(c),(d)

KE = 2 •E(o)

m(o) •c^2•[1/sqrt(1-β^2) -1] =

=2•m(o) •c^2,

1/sqrt(1-β^2) =3,

β = v/c = 0.94.

v =0.94•c =2.82•10^8 m/s.

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