Feynman’s Lectures Exercises 8.2 via the Center-of-Mass Frame

First, calculate the velocity of the CM frame:

${\displaystyle {\mathbf{v}}_{\text{cm}}=\frac{m\mathbf{v}+m\cdot 0}{m+m}=\frac{\mathbf{v}}{2}}$

Now, calculate velocities of particles in the CM frame:

${\displaystyle {\mathbf{v}}_1^{\text{cm}}=\mathbf{v}-{\mathbf{v}}_{\text{cm}}=\frac{\mathbf{v}}{2}}$

${\displaystyle {\mathbf{v}}_2^{\text{cm}}=0-{\mathbf{v}}_{\text{cm}}=-\frac{\mathbf{v}}{2}}$

In the CM frame, particles will have the same speeds after elastic collision, meaning:

${\displaystyle \mathrm{\mid }{\mathbf{u}}_1\mathrm{\mid }=\mathrm{\mid }{\mathbf{u}}_2\mathrm{\mid }=\mathrm{\mid }\mathbf{u}\mathrm{\mid }=\frac{v}{2}}$

The directions of these velocities will be opposite, that is:

${\displaystyle \begin{array}{rl}{\displaystyle {\mathbf{u}}_1}& {\displaystyle =\mathbf{u}}\\ {\displaystyle {\mathbf{u}}_2}& {\displaystyle =-\mathbf{u}}\end{array}}$

Now, let’s move back to the lab frame:

${\displaystyle \begin{array}{rl}{\displaystyle {\mathbf{v}}_1}& {\displaystyle =\mathbf{u}+\frac{\mathbf{v}}{2}}\\ {\displaystyle {\mathbf{v}}_2}& {\displaystyle =-\mathbf{u}+\frac{\mathbf{v}}{2}}\end{array}}$

Let’s calculate the dot product:

${\displaystyle \begin{array}{rl}{\displaystyle {\mathbf{v}}_1\cdot {\mathbf{v}}_2}& {\displaystyle =(\mathbf{u}+\frac{\mathbf{v}}{2})\cdot (-\mathbf{u}+\frac{\mathbf{v}}{2})}\\ {\displaystyle }& {\displaystyle =-\mathrm{\mid }\mathbf{u}{\mathrm{\mid }}^2+\mathbf{u}\cdot \frac{\mathbf{v}}{2}-\mathbf{u}\cdot \frac{\mathbf{v}}{2}+\frac{\mathbf{v}}{2}\cdot \frac{\mathbf{v}}{2}}\\ {\displaystyle }& {\displaystyle =-{\left(\frac{v}{2}\right)}^2+{\left(\frac{v}{2}\right)}^2=0}\end{array}}$

Therefore, the final velocities are orthogonal:

${\displaystyle {\mathbf{v}}_1\perp {\mathbf{v}}_2}$

Previous solution, which doesn’t use the center-of-mass frame, can be found at A more elegant solution can be found in Feynman’s Lectures Exercises 8.2.