Two particles of masses m and 2m start moving simultaneously from...

Let the particles start from the point $(R, 0)$ on a circular track of radius $R$ centered at the origin.

The first particle (mass $m$) moves counter-clockwise. When it covers an angular distance of $45^{\circ}$, its position vector makes an angle of $+45^{\circ}$ with the x-axis. Its velocity vector is tangential to the circular path, making an angle of $45^{\circ} + 90^{\circ} = 135^{\circ}$ with the positive x-axis.

The second particle (mass $2m$) moves clockwise. When it covers an angular distance of $45^{\circ}$, its position vector makes an angle of $-45^{\circ}$ with the x-axis. Its velocity vector is tangential, making an angle of $-45^{\circ} - 90^{\circ} = -135^{\circ}$ with the positive x-axis.

The angle between the two velocity vectors is $135^{\circ} - (-135^{\circ}) = 270^{\circ}$, which is geometrically equivalent to $90^{\circ}$. Thus, the velocity vectors are perpendicular to each other.

The magnitude of the linear momentum of the first particle is $p_1 = mv$.

The magnitude of the linear momentum of the second particle is $p_2 = 2mv$.

Since the two momentum vectors are perpendicular, the magnitude of the total linear momentum of the system is the vector sum:
$P = \sqrt{p_1^2 + p_2^2} = \sqrt{(mv)^2 + (2mv)^2} = \sqrt{m^2v^2 + 4m^2v^2} = \sqrt{5}mv$.

Answer: $\sqrt{5}mv$