A truck of mass `2xx10^(4)` kg travelling at `0.5ms^(-1)` collides with another truck of half its mass moving in the opposite direction with a velocity of `0.4ms^(-1)`. If the trucks couple automatically on collision, calculate the common velocity with which they move.

Answer

To solve the problem, we will use the principle of conservation of linear momentum. The total momentum before the collision must equal the total momentum after the collision since no external forces are acting on the system.

### Step-by-step Solution:

1. **Identify the masses and velocities of the trucks:**
- Mass of Truck 1 (m1) = \(2 \times 10^4\) kg
- Velocity of Truck 1 (u1) = \(0.5\) m/s (moving in the positive direction)
- Mass of Truck 2 (m2) = \(1 \times 10^4\) kg (half of Truck 1)
- Velocity of Truck 2 (u2) = \(-0.4\) m/s (moving in the opposite direction, hence negative)

2. **Calculate the initial momentum of both trucks:**
- Initial momentum of Truck 1 (p1) = \(m1 \times u1 = (2 \times 10^4) \times (0.5) = 1 \times 10^4\) kg·m/s
- Initial momentum of Truck 2 (p2) = \(m2 \times u2 = (1 \times 10^4) \times (-0.4) = -4 \times 10^3\) kg·m/s

3. **Calculate the total initial momentum (P_initial):**
\[ P_{\text{initial}} = p1 + p2 = (1 \times 10^4) + (-4 \times 10^3) = 1 \times 10^4 - 4 \times 10^3 = 6 \times 10^3 \text{ kg·m/s} \]

4. **Determine the total mass after the collision:**
- Total mass (M) = \(m1 + m2 = (2 \times 10^4) + (1 \times 10^4) = 3 \times 10^4\) kg

5. **Use the conservation of momentum to find the common velocity (V):**
- According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{final}} = M \times V \]
\[ 6 \times 10^3 = (3 \times 10^4) \times V \]

6. **Solve for V:**
\[ V = \frac{6 \times 10^3}{3 \times 10^4} = \frac{6}{30} = 0.2 \text{ m/s} \]

7. **Determine the direction of the common velocity:**
- Since the result is positive, the common velocity is in the direction of Truck 1 (the positive direction).

### Final Answer:
The common velocity with which both trucks move after the collision is \(0.2 \, \text{m/s}\) in the positive direction.