Physics, 26.10.2020 17:20 faithcalhoun
A basketball player drops a 0.5-kg basketball vertically so that it is traveling at 5.4 m/s when it reaches the floor. The ball rebounds upward at a speed of 2.6 m/s. Determine (a) the magnitude and direction of the ball's change in momentum and (b) the average net force that the floor exerts on the ball if the collision lasts 0.12 s.
Answers: 1
![\vec F = 33.333\,\hat{j}\,\,\,[N]](/tpl/images/0840/4305/279c2.png)
.
(Eq. 1)
- Mass of the ball, measured in kilograms.
, 
- Velocities of the ball before and after impact.
- Average net force, measured in newtons.
- Collision lapse, measured in seconds. 
(Eq. 1b)
, ![\vec v_{1} = -5.4\,\hat{j}\,\,\,\left[\frac{m}{s} \right]](/tpl/images/0840/4305/4090c.png)
and ![\vec v_{2} = 2.6\,\hat{j}\,\,\,\left[\frac{m}{s} \right]](/tpl/images/0840/4305/eeb11.png)
, the magnitude and direction of the ball's change in momentum is:![\vec F\cdot \Delta t = (0.5\,kg)\cdot \left[2.6\,\hat{j}-(-5.4)\,\hat{j}\right]\,\,\,\left[\frac{m}{s} \right]](/tpl/images/0840/4305/6df6e.png)
![\vec F\cdot \Delta t = 4\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]](/tpl/images/0840/4305/ce392.png)
![\vec F = \frac{4\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]}{0.12\,s}](/tpl/images/0840/4305/c2794.png)
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A basketball player drops a 0.5-kg basketball vertically so that it is traveling at 5.4 m/s when it...
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