A physics teacher throws a 4kg dodgeball 17m/s hitting a 58kg student,who is initially at rest on a skateboard. After hitting the student, the dodgeball rebounds back left with a speed of 5m/s vert Jvert =square unit before collision: Pball=square unit P_(student)=square unit after collision: Pball=square unit P_(student)=square unit v How fast did the student move after the impact? square square v What was the magnitude of the average impact force if the impact lasted 028s?

To solve this problem, we need to apply the principles of conservation of momentum and impulse.<br /><br />### Before Collision:<br /><br />1. **Momentum of the dodgeball (P_ball):**<br /> \[<br /> P_{\text{ball}} = m_{\text{ball}} \times v_{\text{ball}} = 4 \, \text{kg} \times 17 \, \text{m/s} = 68 \, \text{kg} \cdot \text{m/s}<br /> \]<br /><br />2. **Momentum of the student (P_student):**<br /> Since the student is initially at rest:<br /> \[<br /> P_{\text{student}} = m_{\text{student}} \times v_{\text{student}} = 58 \, \text{kg} \times 0 \, \text{m/s} = 0 \, \text{kg} \cdot \text{m/s}<br /> \]<br /><br />### After Collision:<br /><br />1. **Momentum of the dodgeball after rebounding:**<br /> The dodgeball rebounds with a speed of \(5 \, \text{m/s}\) in the opposite direction, so its velocity is \(-5 \, \text{m/s}\):<br /> \[<br /> P_{\text{ball, after}} = 4 \, \text{kg} \times (-5 \, \text{m/s}) = -20 \, \text{kg} \cdot \text{m/s}<br /> \]<br /><br />2. **Conservation of Momentum:**<br /> The total momentum before the collision equals the total momentum after the collision:<br /> \[<br /> P_{\text{total, before}} = P_{\text{total, after}}<br /> \]<br /> \[<br /> 68 \, \text{kg} \cdot \text{m/s} + 0 \, \text{kg} \cdot \text{m/s} = -20 \, \text{kg} \cdot \text{m/s} + P_{\text{student, after}}<br /> \]<br /> Solving for \(P_{\text{student, after}}\):<br /> \[<br /> P_{\text{student, after}} = 68 \, \text{kg} \cdot \text{m/s} + 20 \, \text{kg} \cdot \text{m/s} = 88 \, \text{kg} \cdot \text{m/s}<br /> \]<br /><br />3. **Velocity of the student after impact:**<br /> \[<br /> v_{\text{student, after}} = \frac{P_{\text{student, after}}}{m_{\text{student}}} = \frac{88 \, \text{kg} \cdot \text{m/s}}{58 \, \text{kg}} \approx 1.52 \, \text{m/s}<br /> \]<br /><br />### Impulse and Average Impact Force:<br /><br />1. **Impulse (J):**<br /> Impulse is the change in momentum of the dodgeball:<br /> \[<br /> J = P_{\text{ball, after}} - P_{\text{ball, before}} = -20 \, \text{kg} \cdot \text{m/s} - 68 \, \text{kg} \cdot \text{m/s} = -88 \, \text{kg} \cdot \text{m/s}<br /> \]<br /> The magnitude of the impulse is:<br /> \[<br /> |J| = 88 \, \text{kg} \cdot \text{m/s}<br /> \]<br /><br />2. **Average Impact Force (F_avg):**<br /> Using the impulse-momentum theorem, where impulse equals the average force times the time duration (\(\Delta t\)):<br /> \[<br /> F_{\text{avg}} = \frac{|J|}{\Delta t} = \frac{88 \, \text{kg} \cdot \text{m/s}}{0.28 \, \text{s}} \approx 314.29 \, \text{N}<br /> \]<br /><br />### Summary:<br /><br />- Magnitude of impulse: \(88 \, \text{kg} \cdot \text{m/s}\)<br />- Velocity of the student after impact: \(1.52 \, \text{m/s}\)<br />- Magnitude of the average impact force: \(314.29 \, \text{N}\)