Newtons for 10 seconds.What is the car's final velocity? 9. A408 superball strikes a wall with a velocity of 10m/s that is normal to the wall. It bounces away at a velocity of 7m/s , still normal to the wall. What is the ball's change in momentum? If the bounce lasted 0.1 s, what is the force between the ball and the wall?

To solve this problem, we need to calculate the change in momentum and then use that to find the force.<br /><br />1. **Change in Momentum:**<br /><br /> The initial momentum of the ball before it strikes the wall is:<br /> \[<br /> p_{\text{initial}} = m \cdot v_{\text{initial}}<br /> \]<br /> where \( m = 0.408 \) kg and \( v_{\text{initial}} = 10 \) m/s.<br /> \[<br /> p_{\text{initial}} = 0.408 \times 10 = 4.08 \text{ kg} \cdot \text{m/s}<br /> \]<br /><br /> The final momentum of the ball after it bounces away from the wall is:<br /> \[<br /> p_{\text{final}} = m \cdot v_{\text{final}}<br /> \]<br /> where \( v_{\text{final}} = -7 \) m/s (negative because it is in the opposite direction).<br /> \[<br /> p_{\text{final}} = 0.408 \times (-7) = -2.856 \text{ kg} \cdot \text{m/s}<br /> \]<br /><br /> The change in momentum (\( \Delta p \)) is:<br /> \[<br /> \Delta p = p_{\text{final}} - p_{\text{initial}}<br /> \]<br /> \[<br /> \Delta p = -2.856 - 4.08 = -6.936 \text{ kg} \cdot \text{m/s}<br /> \]<br /><br />2. **Force Calculation:**<br /><br /> The average force exerted on the ball by the wall can be found using the impulse-momentum theorem:<br /> \[<br /> F \cdot \Delta t = \Delta p<br /> \]<br /> where \( \Delta t = 0.1 \) s.<br /><br /> Solving for \( F \):<br /> \[<br /> F = \frac{\Delta p}{\Delta t}<br /> \]<br /> \[<br /> F = \frac{-6.936}{0.1} = -69.36 \text{ N}<br /> \]<br /><br />The negative sign indicates that the force is in the direction opposite to the ball's initial motion.<br /><br />So, the ball's change in momentum is \(-6.936 \text{ kg} \cdot \text{m/s}\), and the force between the ball and the wall is \(69.36 \text{ N}\).