What speed must a particle attain before its kinetic energy is double the value

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We keep in mind that the relativistic KE is given by mc^2 - m_0c^2, with m being the relativistic mass given by m_0/sqrt(1-(v/c)^2)0. We need K_r = 2K_c (relativistic and classical KE, respectively). Equating the two: m_0v^2 = m_0c^2(1/sqrt(1-(v/c)^2) - 1) (v/c)^2 = [1/sqrt(1-[v/c]^2) - 1]. To simplify my equations at this point, I set u = (v/c)^2. u = [1/sqrt(1-u) - 1] u+1 = 1/sqrt(1-u) u^2 + 2u + 1 = 1/(1-u) (u^2 + 2u + 1)(1-u) = 1 -u^3 - u^2 + u = 0 -u(u^2 + u - 1) = 0. Obviously, the relativistic and classical estimates agree when u = 0 --> v = 0. To find the other solutions, we compute the solutions to u^2 + u - 1 = 0, which by the quadratic formula are [-1 +/- sqrt(5)] / 2. Since u is (v/c)^2, we obviously must discard the negative solution (unless you want to throw some serious sci-fi imaginary velocity mess into this), leaving u = (sqrt(5) - 1)/2. Hence v^2 = c^2(sqrt(5) - 1)/2, and so v = sqrt([sqrt(5)-1]/2)c, or about 0.7862c.

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