To solve this problem, let's analyze the given data and use the concept of impulse, which is the change in momentum.

Given Data:

• Initial velocity, $u = 3.0 \, \text{m/s}$ (towards the wall).
• Final velocity, $v = -2.0 \, \text{m/s}$ (away from the wall after rebounding).
• Mass of the ball, $m = 1.0 \, \text{kg}$.
• Contact time with the wall, $t = 0.20 \, \text{s}$.

Solution Steps:

1. Calculate the Change in Momentum: The momentum change $\Delta p$ can be calculated as: $\Delta p = m \cdot (v - u)$ Substitute the values: $\Delta p = 1.0 \cdot (-2.0 - 3.0) = 1.0 \cdot (-5.0) = -5.0 \, \text{kg} \cdot \text{m/s}$

2. Calculate the Impulse: Impulse is equal to the change in momentum: $\text{Impulse} = \Delta p = -5.0 \, \text{kg} \cdot \text{m/s}$

3. Calculate the Force: Force $F$ exerted on the wall is given by the formula: $F = \frac{\text{Impulse}}{t} = \frac{\Delta p}{t}$ Substitute the values: $F = \frac{-5.0}{0.20} = -25.0 \, \text{N}$ The magnitude of the force is therefore: $|F| = 25.0 \, \text{N}$

Answer:

The correct answer is: D. 25 N

Would you like more details on the calculations, or have any further questions?

Related Questions:

1. How is impulse related to momentum and force?
2. What happens to the direction of velocity after a collision with a wall?
3. How would the answer change if the mass of the ball were doubled?
4. Why is the force considered negative in this calculation?
5. How can you calculate the force if the contact time changes?

Tip:

Always keep track of the direction of velocities when dealing with collisions, as it affects the change in momentum and the resulting force direction.