Answer
The final speed is v = 0.41c.
Work Step by Step
The momentum at the final speed is to be two times the initial momentum.
Let the initial state be “1” and the final state be “2”. Use the correct expression for momentum.
$$\frac{p_2}{p_1}=2=\frac{mv_2/(\sqrt{1-v_2^2/c^2})}{ mv_1/(\sqrt{1-v_1^2/c^2})}$$
$$4=\frac{v_2^2/(1-v_2^2/c^2)}{ v_1^2/(1-v_1^2/c^2)}$$
$$4 v_1^2/(1-v_1^2/c^2)= 4 (0.22)^2/(1-0.22^2)=0.20345c^2= v_2^2/(1-v_2^2/c^2)$$
$$v_2^2=\frac{0.20345}{1.20345}c^2$$
$$v_2=0.41c$$
