A ball moving with a momentum of `5kg ms^(-1)` strikes against a wall at an angle of `45^(@)` and is reflected at the same angle Calculate the impulse .

Answer

To solve the problem of calculating the impulse experienced by a ball moving with a momentum of \(5 \, \text{kg m/s}\) when it strikes a wall at an angle of \(45^\circ\) and is reflected at the same angle, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Impulse**:
Impulse is defined as the change in momentum of an object when a force is applied over a period of time. Mathematically, it can be expressed as:
\[ I = \Delta p \]
where \(I\) is the impulse and \(\Delta p\) is the change in momentum.

2. **Identify the Initial Momentum**:
The initial momentum of the ball is given as:
\[ p = 5 \, \text{kg m/s} \]

3. **Resolve Momentum into Components**:
Since the ball strikes the wall at an angle of \(45^\circ\), we need to resolve the momentum into two components: one parallel to the wall (x-direction) and one perpendicular to the wall (y-direction).
- The component of momentum parallel to the wall (x-direction) is:
\[ p_x = p \sin(45^\circ) = 5 \sin(45^\circ) = 5 \cdot \frac{1}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \, \text{kg m/s} \]
- The component of momentum perpendicular to the wall (y-direction) is:
\[ p_y = p \cos(45^\circ) = 5 \cos(45^\circ) = 5 \cdot \frac{1}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \, \text{kg m/s} \]

4. **Determine the Change in Momentum**:
- After the collision, the component of momentum parallel to the wall remains unchanged, while the perpendicular component reverses direction.
- Therefore, the change in momentum in the y-direction (perpendicular to the wall) is:
\[ \Delta p_y = p_y' - p_y = -\frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2} = -5\sqrt{2} \, \text{kg m/s} \]
- The change in momentum in the x-direction is zero since the momentum component parallel to the wall does not change:
\[ \Delta p_x = 0 \]

5. **Calculate the Total Change in Momentum**:
- The total change in momentum (which contributes to the impulse) is only from the y-direction:
\[ \Delta p = \Delta p_x + \Delta p_y = 0 - 5\sqrt{2} = -5\sqrt{2} \, \text{kg m/s} \]

6. **Calculate the Impulse**:
- The impulse experienced by the ball is equal to the change in momentum:
\[ I = \Delta p = -5\sqrt{2} \, \text{kg m/s} \]
- The magnitude of the impulse is:
\[ |I| = 5\sqrt{2} \, \text{kg m/s} \approx 7.07 \, \text{kg m/s} \]

### Final Answer:
The impulse experienced by the ball is approximately \(7.07 \, \text{kg m/s}\) in the direction opposite to the initial y-component of momentum.