To solve the problem, we will apply the principle of conservation of momentum. The total momentum before the explosion must equal the total momentum after the explosion.
### Step-by-Step Solution:
1. **Identify the Initial Momentum:**
The shell is initially at rest, so its initial momentum is:
\[
P_{\text{initial}} = 0 \, \text{kg m/s}
\]
2. **Determine the Momentum of the Two Parts:**
After the explosion, we have two parts of mass 2 kg each.
- The first part moves north with a velocity of 5 m/s.
- The second part moves east with a velocity of 5 m/s.
We can calculate the momentum of each part:
- Momentum of the first part (north):
\[
P_1 = m_1 \cdot v_1 = 2 \, \text{kg} \cdot 5 \, \text{m/s} = 10 \, \text{kg m/s}
\]
- Momentum of the second part (east):
\[
P_2 = m_2 \cdot v_2 = 2 \, \text{kg} \cdot 5 \, \text{m/s} = 10 \, \text{kg m/s}
\]
3. **Calculate the Resultant Momentum of the Two Parts:**
Since the two parts are moving at right angles to each other (north and east), we can find the resultant momentum using the Pythagorean theorem:
\[
P_{\text{resultant}} = \sqrt{P_1^2 + P_2^2} = \sqrt{(10)^2 + (10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{kg m/s}
\]
4. **Determine the Direction of the Resultant Momentum:**
The direction of the resultant momentum is at a 45-degree angle to both the north and east directions, pointing towards the northeast.
5. **Apply Conservation of Momentum:**
According to the conservation of momentum:
\[
P_{\text{initial}} = P_{\text{final}}
\]
Since the initial momentum is zero, the total momentum of all three parts after the explosion must also equal zero:
\[
P_1 + P_2 + P_3 = 0
\]
Therefore, the momentum of the third part (which we denote as \(P_3\)) must be equal in magnitude but opposite in direction to the resultant momentum of the first two parts:
\[
P_3 = -P_{\text{resultant}} = -10\sqrt{2} \, \text{kg m/s}
\]
6. **Calculate the Mass of the Third Part:**
The total mass of the shell is 5 kg, and the two parts have a combined mass of 4 kg (2 kg + 2 kg). Therefore, the mass of the third part is:
\[
m_3 = 5 \, \text{kg} - 4 \, \text{kg} = 1 \, \text{kg}
\]
7. **Find the Velocity of the Third Part:**
Using the momentum formula \(P = m \cdot v\), we can find the velocity of the third part:
\[
P_3 = m_3 \cdot v_3
\]
Substituting the values we have:
\[
-10\sqrt{2} = 1 \cdot v_3 \implies v_3 = -10\sqrt{2} \, \text{m/s}
\]
The negative sign indicates that the direction of the third part is opposite to the resultant momentum of the first two parts, which means it is directed southwest.
### Final Answer:
The velocity of the third part after the explosion is:
\[
v_3 = 10\sqrt{2} \, \text{m/s} \, \text{towards southwest}
\]
