Let's break the problem into two parts and solve each step by step:

Given:

• Mass of the ball, $m = 300 \, \text{g} = 0.300 \, \text{kg}$
• Initial velocity $\vec{v}_1 = -7.0 \, \hat{i} \, \text{m/s}$
• Final velocity after collision $\vec{v}_2 = (-4.0 \, \hat{i} + 2.5 \, \hat{j}) \, \text{m/s}$
• Collision time, $\Delta t = 50 \, \text{ms} = 0.050 \, \text{s}$
• Time after collision, $t = 3.0 \, \text{s}$

(a) Calculate the force exerted on the ball during the impact

The force exerted on the ball can be calculated using Newton's second law and impulse concepts: $\vec{F} = \frac{\Delta \vec{p}}{\Delta t}$ Where $\vec{p}$ is the momentum and $\Delta \vec{p}$ is the change in momentum.

Step 1: Calculate the initial momentum $\vec{p}_1$

$\vec{p}_1 = m \cdot \vec{v}_1 = 0.300 \times (-7.0 \, \hat{i}) = -2.1 \, \hat{i} \, \text{kg·m/s}$

Step 2: Calculate the final momentum $\vec{p}_2$

$\vec{p}_2 = m \cdot \vec{v}_2 = 0.300 \times (-4.0 \, \hat{i} + 2.5 \, \hat{j}) = (-1.2 \, \hat{i} + 0.75 \, \hat{j}) \, \text{kg·m/s}$

Step 3: Calculate the change in momentum $\Delta \vec{p}$

$\Delta \vec{p} = \vec{p}_2 - \vec{p}_1 = (-1.2 \, \hat{i} + 0.75 \, \hat{j}) - (-2.1 \, \hat{i}) = (0.9 \, \hat{i} + 0.75 \, \hat{j}) \, \text{kg·m/s}$

Step 4: Calculate the force $\vec{F}$

$\vec{F} = \frac{\Delta \vec{p}}{\Delta t} = \frac{(0.9 \, \hat{i} + 0.75 \, \hat{j})}{0.050} = (18.0 \, \hat{i} + 15.0 \, \hat{j}) \, \text{N}$

So, the force exerted on the ball during the impact is: $\boxed{\vec{F} = (18.0 \, \hat{i} + 15.0 \, \hat{j}) \, \text{N}}$

(b) Calculate the magnitude of the force opposing the motion of the ball after the collision

After the collision, the ball continues to move for 3.0 seconds. We can assume that a constant opposing force causes the ball to stop. Using kinematics and Newton's second law, we can find this force.

Step 1: Find the deceleration

Assume the ball comes to rest after 3.0 seconds. The final velocity is zero. The initial velocity after the collision is $\vec{v}_2 = (-4.0 \, \hat{i} + 2.5 \, \hat{j}) \, \text{m/s}$.

The magnitude of the initial velocity after the collision: $v_2 = \sqrt{(-4.0)^2 + (2.5)^2} = \sqrt{16 + 6.25} = \sqrt{22.25} \approx 4.72 \, \text{m/s}$

Now, using the equation of motion $v = u + at$, where $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration, and $t$ is the time: $0 = 4.72 + a \cdot 3.0$ Solving for $a$: $a = \frac{-4.72}{3.0} = -1.57 \, \text{m/s}^2$

Step 2: Find the opposing force

The force is given by Newton's second law: $F = m \cdot a = 0.300 \times (-1.57) = -0.471 \, \text{N}$

Thus, the magnitude of the opposing force is: $\boxed{F = 0.471 \, \text{N}}$

Summary:

1. The force exerted on the ball during the impact is $\vec{F} = (18.0 \, \hat{i} + 15.0 \, \hat{j}) \, \text{N}$.
2. The magnitude of the force opposing the motion after the collision is $F = 0.471 \, \text{N}$.

Would you like more details on any part of this solution?

Here are some related questions:

1. What is the significance of impulse in collision problems?
2. How would the result change if the mass of the ball was doubled?
3. Can we calculate the work done on the ball during the impact?
4. What role does the direction of velocity play in calculating forces?
5. How does the duration of collision affect the force?

Tip: In collision problems, always remember that the total momentum before and after the collision is conserved unless an external force acts on the system.