Question

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.

Answer 1

Answer:

a) The magnitude and direction of the ball's change in momentum is 4 kilogram-meter per second upwards.

b) The average net force that the floor exerts on the ball if the collision lasts 0.12 seconds is $\vec F = 33.333\,\hat{j}\,\,\,[N]$.

Explanation:

a) From Principle of Momentum Conservation and Impact Theorem, we represent the situation of the ball as follows:

$m\cdot \vec v_{1}+\vec F \cdot \Delta t = m\cdot \vec v_{2}$ (Eq. 1)

Where:

$m$ - Mass of the ball, measured in kilograms.

$\vec v_{1}$, $\vec v_{2}$ - Velocities of the ball before and after impact.

$\vec F$ - Average net force, measured in newtons.

$\Delta t$ - Collision lapse, measured in seconds.

Now we clear impact within expression:

$\vec F\cdot \Delta t = m\cdot (\vec v_{2}-\vec v_{1})$ (Eq. 1b)

If we know that $m = 0.5\,kg$, $\vec v_{1} = -5.4\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ and $\vec v_{2} = 2.6\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$, the magnitude and direction of the ball's change in momentum is:

$\vec F\cdot \Delta t = (0.5\,kg)\cdot \left[2.6\,\hat{j}-(-5.4)\,\hat{j}\right]\,\,\,\left[\frac{m}{s} \right]$

$\vec F\cdot \Delta t = 4\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The magnitude and direction of the ball's change in momentum is 4 kilogram-meter per second upwards.

b) The average net force is found by dividing result in part a) by given time. That is:

$\vec F = \frac{4\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]}{0.12\,s}$

$\vec F = 33.333\,\hat{j}\,\,\,[N]$

The average net force that the floor exerts on the ball if the collision lasts 0.12 seconds is $\vec F = 33.333\,\hat{j}\,\,\,[N]$.