6. A1.00 kg ball traveling towards a soccer player at a velocity of 5.00m/s rebounds off the soccer player's foot at a velocity of 8.50m/s If the time of contact between the ball and the player's foot was 2.00 s what was the force that the foot applied on the ball?

## Step 1: <br />First, we need to calculate the change in momentum of the ball. The momentum of an object is given by the product of its mass and velocity. The initial momentum of the ball is given by \(p_{i} = m \cdot v_{i}\), where \(m\) is the mass of the ball and \(v_{i}\) is its initial velocity. Similarly, the final momentum of the ball is given by \(p_{f} = m \cdot v_{f}\), where \(v_{f}\) is its final velocity.<br /><br />### \(p_{i} = m \cdot v_{i}\)<br />### \(p_{f} = m \cdot v_{f}\)<br /><br />## Step 2: <br />The change in momentum (\(\Delta p\)) is given by the difference between the final and initial momentum. Since the ball rebounds, the direction of its velocity changes, so we take the final momentum as negative.<br /><br />### \(\Delta p = p_{f} - p_{i}\)<br /><br />## Step 3: <br />Next, we use the impulse-momentum theorem, which states that the impulse experienced by an object is equal to the change in momentum of the object. Impulse is also equal to the force applied on the object times the time for which the force is applied. Therefore, we can write the equation as \(F \cdot t = \Delta p\), where \(F\) is the force applied and \(t\) is the time.<br /><br />### \(F \cdot t = \Delta p\)<br /><br />## Step 4: <br />Finally, we solve for the force \(F\) by rearranging the equation from step 3.<br /><br />### \(F = \frac{\Delta p}{t}\)