Answer
5.6 (this is not the expected classical answer that doubling the speed would quadruple the kinetic energy).
Work Step by Step
Use equation 26–5b to calculate the proton’s kinetic energy. Let subscript 1 represent the original speed (v/c=1/3) and subscript 2 represent the new, higher speed (v/c=2/3).
Find $\gamma$ for this proton.
$$\gamma_1=\frac{1}{\sqrt{1-v_1^2/c^2}}=\frac{1}{\sqrt{1-0.3333^2}}=1.0606$$
$$\gamma_2=\frac{1}{\sqrt{1-v_2^2/c^2}}=\frac{1}{\sqrt{1-0.6667^2}}=1.3417$$
$$\frac{KE_2}{KE_1}=\frac{(\gamma_2 -1)mc^2}{(\gamma_2 -1)mc^2}$$
$$\frac{KE_2}{KE_1}=\frac{0.3417}{0.0606}=5.6$$
