A particle of rest mass \(m_{0}\) travels at a speed \(v=0.20 c\) How fast must the particle travel in order for its momentum to increase to twice its original momentum? a) \(0.40 c\) c) \(0.38 c\) e) \(0.99 c\) b) \(0.10 c\) d) \(0.42 c\)

Using the formula for relativistic momentum, we have: \(p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}\) Where - \(p\) is the momentum of the particle - \(m_0\) is the rest mass of the particle - \(v\) is the speed of the particle - \(c\) is the speed of light

We are given the initial speed of the particle, \(v = 0.20c\). Let's calculate its initial momentum, \(p_0\): \(p_0 = \frac{m_0 (0.20c)}{\sqrt{1 - \frac{(0.20c)^2}{c^2}}} = \frac{0.20m_0c}{\sqrt{1 - 0.04}} = \frac{0.20m_0c}{\sqrt{0.96}}\)

We want to find the speed at which the momentum of the particle becomes twice its initial momentum. So, the equation we need to solve is: \(2p_0 = \frac{m_0v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) Where \(v_f\) is the final speed of the particle.

Now, we can substitute the initial momentum (\(p_0\)) into the equation: \(2\left(\frac{0.20m_0c}{\sqrt{0.96}}\right) = \frac{m_0v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) We can simplify and solve for \(v_f\): \(\frac{0.40m_0c}{\sqrt{0.96}} = \frac{m_0v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) Notice that \(m_0c\) can be canceled from both sides: \(\frac{0.40}{\sqrt{0.96}} = \frac{v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) Now, we can square both sides: \(0.40^2 \cdot 0.96 = (1 - \frac{v_f^2}{c^2})v_f^2\) Simplify and solve for \(v_f\): \(0.1536c^2 = v_f^2 - \frac{v_f^4}{c^2}\) \(v_f^4 - 0.8464c^2v_f^2 + 0.9536c^4 = 0\) This is a quadratic equation in terms of \(v_f^2\). We can use any method to solve it (e.g., factoring, completing the square, or quadratic formula). The solution is: \(v_f^2 = 0.146c^2\) \(v_f = \sqrt{0.146c^2} = 0.382c\)

From the given choices, the speed at which the particle's momentum becomes twice its original value is closest to \(0.38c\). The correct answer is (c) \(0.38c\).