Chapter 2, Problem 1Q

A particle is moving at a speed of less than c/2. If the speed of the particle is doubled, what happens to its momentum?

To determine

The momentum of the particle.

Answer to Problem 1Q

The momentum of the particle increased by a factor of $2\left(\sqrt{\frac{c^2-u^2}{c^2-4u^2}}\right)$.

Explanation of Solution

Write the equation for the relativistic momentum of the [article after increased the speed.

    $p=\left(\frac{mu}{\sqrt{1-\frac{u^2}{c^2}}}\right)$        (I)

Here, $c$ is the speed of light, $p$ is the relativistic momentum and , $u$ is the initial speed.

The final speed is the two times as that of the initial speed which is equal to 2u

Conclusion:

Substitute $2u$ for $u$ in the equation (I) to calculate $p^{\prime }$.

    $\begin{array}{c}{p}^{\prime }=\left(\frac{m2u}{\sqrt{1-\frac{4u^2}{c^2}}}\right)\\ =\frac{m2u}{\sqrt{\frac{c^2-4u^2}{c^2}}}\\ =2mu\left(\sqrt{\frac{c^2}{c^2-4u^2}}\right)\end{array}$        (II)

Here $p^{\prime }$ is the final momentum.

Divide the equation (II) and (I) to calculate the ratio,

$\begin{array}{c}\frac{p^{\prime }}{p}=\frac{2mu\left(\sqrt{\frac{c^2}{c^2-4u^2}}\right)}{\left(\frac{mu}{\sqrt{1-\frac{u^2}{c^2}}}\right)}\\ =2\left(\sqrt{\frac{c^2-u^2}{c^2-4u^2}}\right)\end{array}$

The momentum of the particle increased by a factor of $2\left(\sqrt{\frac{c^2-u^2}{c^2-4u^2}}\right)$.