A ball of mass `20` gram hits a smooth wall at an angle of `45^@)` with a velocity of `15m//s` If ball rebounds at `90^(@)` to the direction of incidence, calculate the impulse received by the ball.

Answer

To solve the problem of calculating the impulse received by the ball when it hits a smooth wall, we can follow these steps:

### Step 1: Understand the Problem
The ball has a mass of 20 grams (which we will convert to kilograms), hits a wall at an angle of 45 degrees with a velocity of 15 m/s, and rebounds at 90 degrees to the direction of incidence.

### Step 2: Convert Mass to Kilograms
The mass of the ball is given as 20 grams. To convert this to kilograms:
\[ m = 20 \text{ grams} = \frac{20}{1000} \text{ kg} = 0.02 \text{ kg} \]

### Step 3: Resolve the Initial Velocity into Components
The initial velocity \( v \) of the ball is 15 m/s at an angle of 45 degrees. We need to resolve this velocity into its x and y components:
- \( v_x = v \cos(45^\circ) = 15 \cdot \frac{1}{\sqrt{2}} = \frac{15}{\sqrt{2}} \)
- \( v_y = v \sin(45^\circ) = 15 \cdot \frac{1}{\sqrt{2}} = \frac{15}{\sqrt{2}} \)

### Step 4: Determine the Final Velocity Components After Rebound
After rebounding at 90 degrees to the direction of incidence, the ball will have:
- Final velocity in the x-direction, \( v'_x = 0 \) (since it moves straight away from the wall)
- Final velocity in the y-direction, \( v'_y = -v_y = -\frac{15}{\sqrt{2}} \) (the negative sign indicates a change in direction)

### Step 5: Calculate the Change in Momentum
The change in momentum (\( \Delta p \)) in the x-direction and y-direction can be calculated as follows:
- Initial momentum in x-direction: \( p_{ix} = m v_x = 0.02 \cdot \frac{15}{\sqrt{2}} \)
- Final momentum in x-direction: \( p_{fx} = m v'_x = 0 \)

Thus, the change in momentum in the x-direction is:
\[ \Delta p_x = p_{fx} - p_{ix} = 0 - (0.02 \cdot \frac{15}{\sqrt{2}}) = -0.02 \cdot \frac{15}{\sqrt{2}} \]

For the y-direction:
- Initial momentum in y-direction: \( p_{iy} = m v_y = 0.02 \cdot \frac{15}{\sqrt{2}} \)
- Final momentum in y-direction: \( p_{fy} = m v'_y = 0.02 \cdot \left(-\frac{15}{\sqrt{2}}\right) = -0.02 \cdot \frac{15}{\sqrt{2}} \)

Thus, the change in momentum in the y-direction is:
\[ \Delta p_y = p_{fy} - p_{iy} = -0.02 \cdot \frac{15}{\sqrt{2}} - (0.02 \cdot \frac{15}{\sqrt{2}}) = -0.04 \cdot \frac{15}{\sqrt{2}} \]

### Step 6: Calculate the Impulse
Impulse is equal to the change in momentum. The total impulse received by the ball is the vector sum of the changes in momentum in both directions:
\[ \text{Impulse} = \Delta p_x + \Delta p_y = -0.02 \cdot \frac{15}{\sqrt{2}} + (-0.04 \cdot \frac{15}{\sqrt{2}}) \]
\[ = -0.06 \cdot \frac{15}{\sqrt{2}} = -\frac{0.9}{\sqrt{2}} \text{ kg m/s} \]

### Step 7: Calculate the Magnitude of Impulse
The magnitude of impulse is:
\[ \text{Impulse} = 0.9 \cdot \frac{1}{\sqrt{2}} \text{ kg m/s} \approx 0.636 \text{ kg m/s} \]

### Final Answer
The impulse received by the ball is approximately \( 0.636 \text{ kg m/s} \).

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