To calculate the magnitude of the applied force, we need to find the impulse delivered during the collision and then use the relationship between impulse and force.

Steps to Solve:

1. Impulse Formula: Impulse ($J$) is the change in momentum of the ball: $J = \Delta p = m \cdot \Delta v$

2. Initial and Final Momentum:

   • Initial velocity: $v_i = 5 \, \text{m/s}$ (toward the wall).
   • Final velocity: $v_f = -3 \, \text{m/s}$ (rebounding from the wall; negative because direction has changed).
   • Change in velocity: $\Delta v = v_f - v_i = -3 - 5 = -8 \, \text{m/s}.$
   • Change in momentum: $\Delta p = m \cdot \Delta v = 2 \cdot (-8) = -16 \, \text{kg·m/s}.$ (Magnitude of impulse: $|J| = 16 \, \text{kg·m/s}$).

3. Relation to Force: The force is related to impulse through the equation: $J = F \cdot \Delta t$ Rearrange to solve for force: $F = \frac{J}{\Delta t}.$

4. Assumption About Time ($\Delta t$): If the time of contact ($\Delta t$) is not given, the applied force cannot be determined directly. For example:

   • If $\Delta t = 0.1 \, \text{s}$, then: $F = \frac{16}{0.1} = 160 \, \text{N}.$

Thus, the magnitude of the force depends on the collision duration. Please provide the collision time to calculate the force, or we can conclude that the impulse magnitude is 16 kg·m/s.

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Would you like more details about impulse and momentum? Here are 5 related questions to expand your understanding:

1. What is the relationship between impulse and force in a practical situation?
2. How does the direction of momentum affect the final calculations?
3. How can we measure the time of contact during a collision in real-life experiments?
4. What happens to the kinetic energy in this collision—was it elastic or inelastic?
5. How would the force change if the mass of the ball doubled?

Tip: Always pay attention to directions when working with vector quantities like velocity and momentum.