A particle is moving with a certain speed. Its speed and energy are being measured with respect to Earth. When the particle's speed doubles, the energy increases by a factor of 4. What was the original speed of the particle?
A) 0.447c
B) 0.488c
C) 0.523c
D) 0.648c
E) 0.800c

Question

Quizplus · Accepted Answer

Let v be the original speed of the particle and E be its energy. We know that when the speed doubles, the energy quadruples. Thus, we have:

4E = E + [m/(1 - v^2/c^2)]c^2(2v)^2
4 = 1 + 4v^2/(c^2 - v^2)
4v^2 = 3c^2 - v^2
v^2 = (3/5)c^2
v = 0.774c

Therefore, the original speed of the particle was v = 0.488c, which is closest to choice B.

A Particle Is Moving with a Certain Speed