The impulse of the net force is approximately \(14.14 \, \text{kg} \cdot \text{m/s}\), and the average net force is approximately \(1414 \, \text{N}\).
The impulse (\(J\)) of a force acting on an object is given by the change in momentum of the object. Mathematically, it is defined as:
\[ J = \Delta p \]
where \( \Delta p \) is the change in momentum, and \( \Delta p = m \cdot \Delta v \), where \( m \) is the mass of the object and \( \Delta v \) is the change in velocity.
1. **Calculate the change in velocity (\( \Delta v \)):**
\[ \Delta v = v_f - v_i \]
\[ \Delta v = (v_f \cdot \cos(\theta), v_f \cdot \sin(\theta)) - (v_{i_x}, v_{i_y}) \]
\[ \Delta v = (30 \cdot \cos(45^\circ) - (-20), 30 \cdot \sin(45^\circ) - 0) \]
\[ \Delta v = (30 \cdot \frac{\sqrt{2}}{2} + 20, 30 \cdot \frac{\sqrt{2}}{2}) \]
\[ \Delta v = (20\sqrt{2} + 20, 15\sqrt{2}) \]
2. **Calculate the change in momentum (\( \Delta p \)):**
\[ \Delta p = m \cdot \Delta v \]
\[ \Delta p = 0.40 \cdot (20\sqrt{2} + 20, 15\sqrt{2}) \]
\[ \Delta p = (8\sqrt{2} + 8, 6\sqrt{2}) \]
3. **Calculate the impulse (\( J \)):**
\[ J = \sqrt{(8\sqrt{2} + 8)^2 + (6\sqrt{2})^2} \]
\[ J \approx \sqrt{128 + 72} \]
\[ J \approx \sqrt{200} \]
\[ J \approx 14.14 \, \text{kg} \cdot \text{m/s} \]
4. **Calculate the average net force (\( F_{\text{avg}} \)):**
\[ F_{\text{avg}} = \frac{J}{\Delta t} \]
\[ F_{\text{avg}} = \frac{14.14}{0.010} \]
\[ F_{\text{avg}} \approx 1414 \, \text{N} \]
Therefore, the impulse of the net force is approximately \(14.14 \, \text{kg} \cdot \text{m/s}\), and the average net force is approximately \(1414 \, \text{N}\).