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?

Question

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?

Answer 1

To solve this problem, we need to apply the principles of conservation of momentum and impulse.### Before Collision:1. **Momentum of the dodgeball (P_ball):**

2. **Momentum of the student (P_student):** Since the student is initially at rest:

### After Collision:1. **Momentum of the dodgeball after rebounding:** The dodgeball rebounds with a speed of

in the opposite direction, so its velocity is

:

2. **Conservation of Momentum:** The total momentum before the collision equals the total momentum after the collision:

Solving for

:

3. **Velocity of the student after impact:**

### Impulse and Average Impact Force:1. **Impulse (J):** Impulse is the change in momentum of the dodgeball:

The magnitude of the impulse is:

2. **Average Impact Force (F_avg):** Using the impulse-momentum theorem, where impulse equals the average force times the time duration (

):

### Summary:- Magnitude of impulse:

- Velocity of the student after impact:

- Magnitude of the average impact force:

Question

Solution

Answer