Two particles of masses $m_1$ and $m_2$ in projectile motion have velocities $\vec{u}_1$ and $\vec{u}_1$ respectively at time $t = 0$. They collide at time $t_0$. Their velocities become $\vec{u}_2$ and $\vec{u}_2'$ at time $2t_0$ while still moving in air. The value of $|(m_1\vec{u}_1' + \vec{u}_2') - (m_1\vec{u}_1 + m_2\vec{u}_2')|$ is

For the two-particle system, during motion in air the only external force is gravity, so the total momentum changes according to impulse of gravity. At $t=0$, total momentum is $$\vec{P}_0=m_1\vec{u}_1+m_2\vec{u}_2$$ At $t=2t_0$, total momentum is $$\vec{P}_{2t_0}=m_1\vec{u}_1'+m_2\vec{u}_2'$$ Hence, $$\vec{P}_{2t_0}-\vec{P}_0=(m_1+m_2)\vec{g}\,(2t_0)$$ Its magnitude is $$\left|\vec{P}_{2t_0}-\vec{P}_0\right|=2(m_1+m_2)gt_0$$ The given expression is exactly the magnitude of final total momentum minus initial total momentum, so it equals this value. On checking the options, this matches option $C$.