Two particles of masses $m_1$ and $m_2$ in projectile motion have velocities $v_1$ and $v_2$ respectively at time $t = 0$. They collide at time $t_o$. Their velocities become $v_1'$ and $v_2'$ at time $2t_o$ while still moving in air. The value of $\left|(m_1 v_1' + m_2 v_2') - (m_1 v_1 + m_2 v_2)\right|$ is

Step 1: Define the system and total momentum.

Consider the system consisting of both particles of masses $m_1$ and $m_2$.

$$\vec{P}(t) = m_1\vec{v}_1(t) + m_2\vec{v}_2(t)$$

Step 2: Identify the forces acting on the system.

The forces during the collision at $t = t_o$ are internal forces, which do not change the total momentum of the system.

$$\vec{F}_{\text{int}} = \vec{0}$$

The only external force acting on the system throughout the motion in the air is gravity.

$$\vec{F}_{\text{ext}} = (m_1 + m_2)\vec{g}$$

Step 3: Calculate the change in total momentum.

The change in total momentum from $t = 0$ to $t = 2t_o$ is equal to the impulse of the external gravitational force over this time interval.

$$\Delta\vec{P} = \vec{P}(2t_o) - \vec{P}(0)$$ $$\Delta\vec{P} = (m_1 + m_2)\vec{g}(2t_o)$$

Step 4: Find the magnitude of the change in momentum.

Take the magnitude of the momentum change vector.

$$\left|\left(m_1\vec{v}_1' + m_2\vec{v}_2'\right) - \left(m_1\vec{v}_1 + m_2\vec{v}_2\right)\right| = \left|\Delta\vec{P}\right|$$ $$\left|\Delta\vec{P}\right| = 2(m_1 + m_2)g t_o$$

Answer: $2(m_1 + m_2)g t_o$