If the kinetic energy of a particle is 3.00 times its rest energy, the speed of
ID: 1384958 • Letter: I
Question
If the kinetic energy of a particle is 3.00 times its rest energy, the speed of the particle is closest to which of the following values? I know the answer is B. [.968c]... I just need to know how the problem is broken down and what variables to put where when I finally have derived the equation to:
4 { 1 - ( v / c ) ^ 2 } ^ (1/2) = 1 ........ I am unsure of how I get to the answer of B. If the previous equation is right it should be pretty easy, if not then please the correct equation with values correlating to variables thank you!
A) .950c
B) .968c
C) .976c
D) .984c
Explanation / Answer
KE = total energy - rest mass energy
KE = mc^2 - moc^2
m = mo/sqrt(1-v^2/c^2)
KE = mo*c^2/sqrt(1-v^/c^2) - mo*c^2
KE = mo*c^2*[ (1/sqrt(1-v^2/c^2)) - 1 ]
given KE = 3*mo*c^2
mo*c^2*[ 1/sqrt(1-v^2/c^2) - 1 ] = 3*mo*c^2
1/sqrt(1-v^2/c^2) - 1 = 3
sqrt(1-v^2/c^2) = 1/4
1- v^2/c^2 = 1/16
v^2/c^2 = 0.9375
v = 0.968 c <-----answer