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

Here's the solution:<br /><br />**1. Impulse:**<br /><br />The impulse ($\vert J\vert$) 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:<br /><br />* Initial momentum of the ball: $p_{ball,i} = (7 kg)(20 m/s) = 140 kg \cdot m/s$<br />* Final momentum of the ball: $p_{ball,f} = (7 kg)(-6 m/s) = -42 kg \cdot m/s$ (negative since it rebounds to the left)<br /><br />Therefore, the impulse is:<br /><br />$\vert J\vert = |p_{ball,f} - p_{ball,i}| = |-42 kg \cdot m/s - 140 kg \cdot m/s| = |-182 kg \cdot m/s| = 182 kg \cdot m/s$<br /><br />**Before collision:**<br /><br />* $P_{ball} = 140 \, kg \cdot m/s$<br />* $P_{student} = 0 \, kg \cdot m/s$ (at rest)<br /><br />**After collision:**<br /><br />* $P_{ball} = -42 \, kg \cdot m/s$<br />* $P_{student} = 182 \, kg \cdot m/s$ (due to conservation of momentum - see below)<br /><br />**2. Student's Speed After Impact:**<br /><br />By the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision.<br /><br />$P_{ball,i} + P_{student,i} = P_{ball,f} + P_{student,f}$<br /><br />$140 kg \cdot m/s + 0 = -42 kg \cdot m/s + (60 kg)v_{student}$<br /><br />$182 kg \cdot m/s = (60 kg)v_{student}$<br /><br />$v_{student} = \frac{182 kg \cdot m/s}{60 kg} = 3.03 \, m/s$<br /><br />**3. Magnitude of Average Impact Force:**<br /><br />The impulse is also equal to the average force multiplied by the time of impact:<br /><br />$\vert J\vert = F_{avg} \cdot \Delta t$<br /><br />$182 kg \cdot m/s = F_{avg} \cdot 0.49 s$<br /><br />$F_{avg} = \frac{182 kg \cdot m/s}{0.49 s} = 371.43 \, N$<br /><br /><br />**Final Answers:**<br /><br />* $\vert J\vert = 182 \, kg \cdot m/s$<br />* Before collision:<br /> * $P_{ball} = 140 \, kg \cdot m/s$<br /> * $P_{student} = 0 \, kg \cdot m/s$<br />* After collision:<br /> * $P_{ball} = -42 \, kg \cdot m/s$<br /> * $P_{student} = 182 \, kg \cdot m/s$<br />* Student's speed after impact: $3.03 \, m/s$<br />* Magnitude of average impact force: $371.43 \, N$<br />