Let $\mathbf{p}$ be the projection of $\mathbf{v}$ onto $\mathbf{w},$ and let $\mathbf{q}$ be the projection of $\mathbf{p}$ onto $\mathbf{v}.$  If $\frac{\|\mathbf{p}\|}{\|\mathbf{v}\|} = \frac{5}{7},$ then find $\frac{\|\mathbf{q}\|}{\|\mathbf{v}\|}.$
This problem involves vectors, projections, and norms (lengths). I recall that the projection of $\mathbf{v}$ onto $\mathbf{w}$ is given by $\mathbf{p} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \mathbf{w}.$ Similarly, the projection of $\mathbf{p}$ onto $\mathbf{v}$ is given by $\mathbf{q} = \frac{\mathbf{p} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.$ I also recall that the norm of a vector is related to the dot product by $\|\mathbf{v}\|^2 = \mathbf{v} \cdot \mathbf{v}.$ I notice that $\mathbf{p} \cdot \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} (\mathbf{w} \cdot \mathbf{v}) = \frac{(\mathbf{v} \cdot \mathbf{w})^2}{\mathbf{w} \cdot \mathbf{w}}.$ I also notice that $\mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\| \|\mathbf{w}\| \cos \theta,$ where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{w}.$ Plugging this into the expression for $\|\mathbf{q}\|,$ I get $\|\mathbf{q}\| = \frac{\|\mathbf{v}\|^2 \cos^2 \theta}{\|\mathbf{v}\|^2} \|\mathbf{v}\| = \|\mathbf{v}\| \cos^2 \theta.$ Now I need to find $\cos \theta.$ I can use the fact that $\frac{\|\mathbf{p}\|}{\|\mathbf{v}\|} = \frac{5}{7}$ to do this. I know that $\|\mathbf{p}\| = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \|\mathbf{w}\| = \frac{\|\mathbf{v}\| \|\mathbf{w}\| \cos \theta}{\|\mathbf{w}\|} = \|\mathbf{v}\| \cos \theta.$ Therefore, $\frac{\|\mathbf{p}\|}{\|\mathbf{v}\|} = \frac{5}{7}$ implies that $\cos \theta = \frac{5}{7}.$ Finally, I can substitute this into the expression for $\|\mathbf{q}\|,$ and get $\|\mathbf{q}\| = \|\mathbf{v}\| \cos^2 \theta = \|\mathbf{v}\| \left(\frac{5}{7}\right)^2 = \frac{25}{49} \|\mathbf{v}\|.$ Hence, $\frac{\|\mathbf{q}\|}{\|\mathbf{v}\|} = \frac{25}{49}.$ 