Question
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
Solution
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Howard
Professional · Tutor for 6 years
Answer
Here's the solution:1. **Impulse Calculation:**Impulse is the change in momentum. For the dodgeball:* Initial momentum: (6 kg)(21 m/s) = 126 kg m/s* Final momentum: (6 kg)(-4 m/s) = -24 kg m/s (Note the negative sign because the ball rebounds in the opposite direction)Therefore, the impulse experienced by the ball is:|J| = |-24 kg m/s - 126 kg m/s| = |-150 kg m/s| = 150 kg m/s**Answer:** |J| = 150 kg m/s2. **Momentum after collision (ball):**As calculated above, the final momentum of the ball is -24 kg m/s.**Answer:** *P*ball = -24 kg m/s3. **Student's speed after impact:**The principle of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision.* Total momentum before collision: (6 kg)(21 m/s) + (60 kg)(0 m/s) = 126 kg m/s* Total momentum after collision: (6 kg)(-4 m/s) + (60 kg)(v) , where 'v' is the student's velocityTherefore:126 kg m/s = -24 kg m/s + 60 kg * v150 kg m/s = 60 kg * vv = 2.5 m/s**Answer:** 2.5 m/s4. **Magnitude of average impact force:**Impulse is also equal to the average force multiplied by the time of impact:|J| = Favg * t150 kg m/s = Favg * 0.29 sFavg = 150 kg m/s / 0.29 s Favg ≈ 517.24 N**Answer:** 517 N (The unit is Newtons, represented by 'N')