A physics teacher throws a 6kg dodgeball 21m/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 4m/s vert Jvert =square kgm/s before collision: P_(student)=0 kgm/s after collision: P_(ball)=square kgm/s P_(student)=150kgm/s How fast did the student move after the impact? m/s What was the magnitude of the average Impact force if the impact lasted 0.29s? 517 square v

Here's the solution:<br /><br />1. **Impulse Calculation:**<br /><br />Impulse is the change in momentum. For the dodgeball:<br /><br />* Initial momentum: (6 kg)(21 m/s) = 126 kg m/s<br />* Final momentum: (6 kg)(-4 m/s) = -24 kg m/s (Note the negative sign because the ball rebounds in the opposite direction)<br /><br />Therefore, the impulse experienced by the ball is:<br /><br />|J| = |-24 kg m/s - 126 kg m/s| = |-150 kg m/s| = 150 kg m/s<br /><br />**Answer:** |J| = 150 kg m/s<br /><br />2. **Momentum after collision (ball):**<br /><br />As calculated above, the final momentum of the ball is -24 kg m/s.<br /><br />**Answer:** *P*_ball = -24 kg m/s<br /><br />3. **Student's speed after impact:**<br /><br />The principle of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision.<br /><br />* Total momentum before collision: (6 kg)(21 m/s) + (60 kg)(0 m/s) = 126 kg m/s<br />* Total momentum after collision: (6 kg)(-4 m/s) + (60 kg)(v) , where 'v' is the student's velocity<br /><br />Therefore:<br /><br />126 kg m/s = -24 kg m/s + 60 kg * v<br />150 kg m/s = 60 kg * v<br />v = 2.5 m/s<br /><br />**Answer:** 2.5 m/s<br /><br />4. **Magnitude of average impact force:**<br /><br />Impulse is also equal to the average force multiplied by the time of impact:<br /><br />|J| = F_avg * t<br /><br />150 kg m/s = F_avg * 0.29 s<br />F_avg = 150 kg m/s / 0.29 s <br />F_avg ≈ 517.24 N<br /><br />**Answer:** 517 N (The unit is Newtons, represented by 'N')<br />