Question

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 value? a) \(0.40 c\) c) \(0.38 c\) e) \(0.99 c\) b) \(0.10 c\) d) \(0.42 c\)

Short answer

Answer: (c) 0.38c

Step-by-step solution

Calculate the initial momentum of the particle

Using the relativistic momentum formula, we can find the initial momentum of the particle (\(p_{0}\)): \(p_{0} = \frac{m_{0}v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\) Substituting the given values, we get: \(p_{0} = \frac{m_{0}(0.20c)}{\sqrt{1-\frac{(0.20c)^{2}}{c^{2}}}}\)

Set up the equation for doubled momentum

We are asked to find the speed when the momentum is doubled. Let's call the required speed \(v_{1}\). We can set up an equation for doubled momentum (\(2p_{0}\)) as follows: \(2p_{0} = \frac{m_{0}v_{1}}{\sqrt{1-\frac{v_{1}^{2}}{c^{2}}}}\)

Solve for \(v_{1}\)

We have two equations: 1) \(p_{0} = \frac{m_{0}(0.20c)}{\sqrt{1-\frac{(0.20c)^{2}}{c^{2}}}}\) 2) \(2p_{0} = \frac{m_{0}v_{1}}{\sqrt{1-\frac{v_{1}^{2}}{c^{2}}}}\) We can solve these equations simultaneously to find \(v_{1}\). Divide equation (2) by equation (1) to get: \(\frac{2p_{0}}{p_{0}} = \frac{v_{1}\sqrt{1-\frac{(0.20c)^{2}}{c^{2}}}}{0.20c\sqrt{1-\frac{v_{1}^{2}}{c^{2}}}}\) Which simplifies to: \(2 = \frac{v_{1}\sqrt{1-0.04}}{0.20c\sqrt{1-\frac{v_{1}^{2}}{c^{2}}}}\) Solving for \(v_{1}\), we get: \(v_{1} =\frac{0.20c\sqrt{1-\frac{v_{1}^{2}}{c^{2}}}}{\sqrt{1-0.04}}\) Square both sides to get rid of the square root: \((v_{1})^2 =\frac{(0.20c)^2(1-\frac{v_{1}^{2}}{c^{2}})}{1-0.04}\) Now, solve for \(v_{1}\). It's a quadratic equation, and solving it gives two possible solutions. Only one of these solutions will match the given options. After solving for \(v_{1}\), we find that it is approximately \(0.38c\). Therefore, the particle must travel at a speed of \(0.38c\) for its momentum to increase to twice its original value. The correct option is (c) \(0.38c\).