Question

The kinetic energy for a certain particle is 1.5 times as large as its rest energy.

A.) what is the speed of this particle?

B.) by what factor do we need to increase the speed of this particle to triple its kinetic energy?

Answer 1

A] We know, KE = m0c^2 *(1/sqrt(1- v^2/c^2) -1)

Here, 1.5 = (1/sqrt(1- v^2/c^2) -1)

2.5 = 1/sqrt(1- v^2/c^2)

sqrt(1- v^2/c^2) = 0.4

1 - v^2/c^2 = 0.16

1 - 0.16= v^2/c^2

v = sqrt (0.84 ) c

= 0.9165c answer

B] KE = m0c^2 *(1/sqrt(1- v^2/c^2) -1)

1.5*3* m0c^2 =  m0c^2 *(1/sqrt(1- v^2/c^2) -1)

4.5 = (1/sqrt(1- v^2/c^2) -1)

5.5 = 1/sqrt(1- v^2/c^2)

sqrt(1- v^2/c^2) = 1/5.5

1- v^2/c^2 = ( 1/5.5)^2

  v^2/c^2 = 1 -  ( 1/5.5)^2 = 0.9669

V = sqrt(0.9669) c = 0.9833 c

increase in v = 0.9833/ 0.9165

= 1.073 times answer