91Ó°ÊÓ

Suppose an object is moving with the velocity$u\text{'}$ with respect to ³§â€™ 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

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

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 ³§â€™ 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 ³§â€™ 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 ³§â€™ frame. The speed of particle 1 $u{\text{'}}_2$in ³§â€™ 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,

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

As$vâ\copyright ½c$, the value of v is $\left(2-\sqrt{3}\right)c=0.268c≈0.27c$.

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