Two particles of masses `m_(1)` and `m_(2)` in projectile motion have velocities `vec(v)_(1)` and `vec(v)_(2)` , respectively , at time `t = 0`. They collide at time `t_(0)`. Their velocities become `vec(v')_(1)` and `vec(v')_(2)` at time ` 2 t_(0)` while still moving in air. The value of `|(m_(1) vec(v')_(1) + m_(2) vec(v')_(2)) - (m_(1) vec(v)_(1) + m_(2) vec(v)_(2))|`

Options

A. zero
B. `(m_(1)+m_(2))t_(0)`
C. `1/2(m_(1)+m_(2))t_(0)`
D. `2(m_(1)+m_(2))t_(0)`

Explanation

To solve the problem, we need to analyze the momentum of the two particles before and after the collision and apply the principle of conservation of momentum.

### Step-by-Step Solution:

1. **Identify Initial Momentum:**
At time \( t = 0 \), the initial momentum \( \vec{p}_1 \) of the system (two particles) is given by:
\[ \vec{p}_1 = m_1 \vec{v}_1 + m_2 \vec{v}_2 \]

2. **Identify Final Momentum:**
At time \( t = 2t_0 \), after the collision, the final momentum \( \vec{p}_2 \) of the system is:
\[ \vec{p}_2 = m_1 \vec{v}'_1 + m_2 \vec{v}'_2 \]

3. **Calculate the Change in Momentum:**
The change in momentum \( \Delta \vec{p} \) is given by:
\[ \Delta \vec{p} = \vec{p}_2 - \vec{p}_1 = (m_1 \vec{v}'_1 + m_2 \vec{v}'_2) - (m_1 \vec{v}_1 + m_2 \vec{v}_2) \]

4. **Magnitude of Change in Momentum:**
We need to find the magnitude of this change in momentum:
\[ |\Delta \vec{p}| = |(m_1 \vec{v}'_1 + m_2 \vec{v}'_2) - (m_1 \vec{v}_1 + m_2 \vec{v}_2)| \]

5. **Consider the Impulse-Momentum Theorem:**
The change in momentum is equal to the impulse applied to the system. Since the only external force acting on the system is gravity, we can express the impulse as:
\[ \Delta \vec{p} = \vec{F}_{\text{external}} \cdot \Delta t \]
where \( \Delta t = 2t_0 \).

6. **Calculate the External Force:**
The gravitational force acting on the system is:
\[ \vec{F}_{\text{external}} = (m_1 + m_2)g \]
Thus, the impulse becomes:
\[ \Delta \vec{p} = (m_1 + m_2)g \cdot (2t_0) \]

7. **Final Result:**
Therefore, the magnitude of the change in momentum is:
\[ |\Delta \vec{p}| = 2(m_1 + m_2)gt_0 \]

### Final Answer:
The value of \( |\Delta \vec{p}| \) is:
\[ 2(m_1 + m_2)gt_0 \]