A particle moves with a constant speed v in a circular path. When...

Let the initial velocity be $\vec{v}_1$ and final velocity be $\vec{v}_2$.

Since the particle moves with a constant speed $v$, the magnitude of velocity remains constant. Therefore, the change in magnitude of velocity is:
$\Delta |\vec{v}| = |\vec{v}_2| - |\vec{v}_1| = v - v = 0$

The magnitude of change in velocity is given by the vector subtraction of $\vec{v}_1$ from $\vec{v}_2$:
$|\Delta \vec{v}| = |\vec{v}_2 - \vec{v}_1| = \sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos\theta}$

Given $\theta = 60^{\circ}$ and $v_1 = v_2 = v$:
$|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2(v)(v)\cos 60^{\circ}}$
$|\Delta \vec{v}| = \sqrt{2v^2 - 2v^2 \left(\dfrac{1}{2}\right)} = \sqrt{v^2} = v$

Thus, the magnitude of change in velocity is $v$, and the change in magnitude of velocity is $0$.

Answer: $v, 0$