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

First, calculate the velocity of the CM frame:

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

Now, calculate velocities of particles in the CM frame:

$$\mathbf{v}_1^{\text{cm}} = \mathbf{v} - \mathbf{v}_{\text{cm}} = \frac{\mathbf{v}}{2}$$ $$\mathbf{v}_2^{\text{cm}} = 0 - \mathbf{v}_{\text{cm}} = -\frac{\mathbf{v}}{2}$$

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

$$|\mathbf{u}_1| = |\mathbf{u}_2| = |\mathbf{u}| = \frac{v}{2}$$

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

$$\begin{aligned} \mathbf{u}_1 &= \mathbf{u} \\ \mathbf{u}_2 &= -\mathbf{u} \end{aligned}$$

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

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

Let’s calculate the dot product:

$$\begin{aligned} \mathbf{v}_1 \cdot \mathbf{v}_2 &= \left( \mathbf{u} + \frac{\mathbf{v}}{2} \right) \cdot \left( -\mathbf{u} + \frac{\mathbf{v}}{2} \right) \\[10pt] &= -|\mathbf{u}|^2 + \mathbf{u} \cdot \frac{\mathbf{v}}{2} - \mathbf{u} \cdot \frac{\mathbf{v}}{2} + \frac{\mathbf{v}}{2} \cdot \frac{\mathbf{v}}{2} \\[10pt] &= -\left( \frac{v}{2} \right)^2 + \left( \frac{v}{2} \right)^2 = 0 \end{aligned}$$

Therefore, the final velocities are orthogonal:

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

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.