1. Calculate the momentum of the baseball as it strikes the wal Diagram \( A \). 2. Calculate the momentum of the baseball as it leaves the wall Diagram \( A \). 3. Calculate the change in momentum during the collision in Diagram \( A \). 4. Calculate the force exerted by the wall on the baseball in Diagram \( A \) and in Diagram \( B \), if each collision lasts \( 0,1 \mathrm{~s} \).

To assist you effectively with the momentum and force calculations for the baseball collision described in **Diagram A** and **Diagram B**, I'll outline the general approach and the necessary formulas. However, to provide specific numerical answers, I'll need additional information such as the mass of the baseball and its velocity before and after the collision. If you have these details, please provide them. In the meantime, here's how you can approach each part of the problem: --- ### **1. Calculating the Momentum of the Baseball as It Strikes the Wall (Diagram A)** **Momentum (\( p \))** is calculated using the formula: \[ p = m \times v \] - **\( m \)**: Mass of the baseball (in kilograms, kg) - **\( v \)**: Velocity of the baseball as it strikes the wall (in meters per second, m/s) **Example Calculation:** If the baseball has a mass of 0.15 kg and is traveling at 30 m/s towards the wall: \[ p_{\text{before}} = 0.15\, \text{kg} \times 30\, \text{m/s} = 4.5\, \text{kg·m/s} \] --- ### **2. Calculating the Momentum of the Baseball as It Leaves the Wall (Diagram A)** Assuming the baseball rebounds off the wall, its velocity will have the same magnitude but opposite direction. Therefore, if the velocity after collision is \( v' = -v \): \[ p_{\text{after}} = m \times v' = m \times (-v) = -m \times v \] **Using the Example:** \[ p_{\text{after}} = 0.15\, \text{kg} \times (-30\, \text{m/s}) = -4.5\, \text{kg·m/s} \] *The negative sign indicates the direction of momentum is opposite after the collision.* --- ### **3. Calculating the Change in Momentum During the Collision in Diagram A** **Change in Momentum (\( \Delta p \))** is given by: \[ \Delta p = p_{\text{after}} - p_{\text{before}} \] **Using the Example:** \[ \Delta p = (-4.5\, \text{kg·m/s}) - (4.5\, \text{kg·m/s}) = -9.0\, \text{kg·m/s} \] *The negative sign indicates that the momentum has reversed direction.* --- ### **4. Calculating the Force Exerted by the Wall on the Baseball (Diagrams A & B)** **Force (\( F \))** can be determined using the impulse-momentum theorem: \[ F = \frac{\Delta p}{\Delta t} \] - **\( \Delta p \)**: Change in momentum - **\( \Delta t \)**: Collision time (given as 0.1 seconds) **Using the Example:** \[ F = \frac{-9.0\, \text{kg·m/s}}{0.1\, \text{s}} = -90\, \text{N} \] *The negative sign indicates that the force exerted by the wall is in the opposite direction of the baseball's initial motion.* --- ### **Applying to Diagram B** Since **Diagram B** isn't described, the force calculation would follow the same principles. If Diagram B involves different parameters (e.g., a different mass, velocity, or collision time), substitute those values into the formulas provided above. --- ### **Summary of Formulas** 1. **Momentum Before Collision:** \[ p_{\text{before}} = m \times v \] 2. **Momentum After Collision:** \[ p_{\text{after}} = m \times v' \] 3. **Change in Momentum:** \[ \Delta p = p_{\text{after}} - p_{\text{before}} \] 4. **Force Exerted by the Wall:** \[ F = \frac{\Delta p}{\Delta t} \] --- ### **Additional Information Needed** To provide precise numerical answers, please supply the following details: - **Mass of the Baseball (\( m \))** - **Velocity Before Collision (\( v \))** - **Velocity After Collision (\( v' \))** *(if different from the example)* - **Any Specifics About Diagram B** Feel free to share these details, and I can help you work through the calculations step by step!