Question

A particle with mass m has speed c/2 relative to inertial frame S. The particle collides with an identical particle at rest relative to frame S. Relative to S, what is the speed of a frame S’ in which the total momentum of these particles is zero? This frame is called the center of the momentum frame.

Short answer

The speed of the center of the momentum frame is 0.27c relative to the S frame.

Step-by-step solution

Relativistic velocity addition

Suppose an object is moving with the velocity$u\text{'}$ with respect to S’ frame, which is moving with the velocitydata-custom-editor="chemistry" $\mathrm{v}$ with respect to frame Sthen the velocity of the object$u$ with respect to S frame is

$u=\frac{u\text{'}+v}{1+\frac{u\text{'}v}{c^2}}$

Relativistic velocity expression for the given scenario.

Let’s consider the stationary particle in S fame as particle 1. Particle 2 is moving at speed $\left(c/2\right)$towards particle 1. Let’s consider another frame S’ in which the system’s total momentum is zero. This can happen when both particles move at the same speed in the opposite direction. Let’s say S’ frame is moving at a speed relative to the S frame. As particle 1 was stationary in the S frame, it will have speed $u{\text{'}}_1=-v$in S’ frame. The speed of particle 1 $u{\text{'}}_2$in S’ frame is

${u\text{'}}_2=\frac{u_2-v}{1-\frac{u_{2}v}{c^2}}$

$-u_1^{\text{'}}=\frac{{\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-V}{1-\left({\displaystyle \frac{c}{2}}\right)\left({\displaystyle \frac{V}{c^2}}\right)}$

$$\begin{gathered} -\left(-V\right)=\frac{{\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-V}{1-\left({\displaystyle \frac{V}{2c}}\right)} \\ \frac{{\displaystyle \frac{{\displaystyle \frac{c-2V}{2}}}{2c-V}}}{2c}=V \end{gathered}$$

Rearranging the equation

$$\begin{gathered} \frac{c\left(c-2v\right)}{2c-v}=v \\ {c}^2-2cv=2cv-v^2 \\ {v}^2-4cv+c^2=0 \end{gathered}$$

Solving the above equation using the quadratic formula,

$$\begin{gathered} v=\frac{-\left(-4c\right)\pm \sqrt{{\left(-4c\right)}^2-4\left(c^2\right)}}{2} \\ =\frac{4c\pm 2\sqrt{3}c}{2} \\ =c\left(2\pm \sqrt{3}\right) \end{gathered}$$

As$v⩽c$, the value of v is $\left(2-\sqrt{3}\right)c=0.268c\approx 0.27c$.

Hence, the speed of the center of the momentum frame is 0.27c relative to the S frame.