A physics teacher throws a 7kg dodgeball 20m/s hitting a 60kg student who is initially at rest on a skateboard. After hitting the student, the dodgeball rebounds back left with a speed of 6m/s vert Jvert =square unit before collision: Pball=square unit vee P_(student)=square unit vee after collision: Pball=square unit vee 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 0 .49s? square square

Question

A physics teacher throws a 7kg dodgeball 20m/s hitting a 60kg student who is initially at rest on a skateboard. After hitting the student, the dodgeball rebounds back left with a speed of 6m/s vert Jvert =square unit before collision: Pball=square unit vee P_(student)=square unit vee after collision: Pball=square unit vee 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 0 .49s? square square

Answer 1

Here's the solution:**1. Impulse:**The impulse (

) is the change in momentum of the dodgeball. Since momentum is a vector, we need to consider the direction. Taking "to the right" as positive:* Initial momentum of the ball:

* Final momentum of the ball:

(negative since it rebounds to the left)Therefore, the impulse is:

**Before collision:***

*

(at rest)**After collision:***

*

(due to conservation of momentum - see below)**2. Student's Speed After Impact:**By the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision.

**3. Magnitude of Average Impact Force:**The impulse is also equal to the average force multiplied by the time of impact:

**Final Answers:***

* Before collision: *

*

* After collision: *

*

* Student's speed after impact:

* Magnitude of average impact force:

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