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Expert-verified Found in: Page 284 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # Review: After a ${\mathbf{0}}{\mathbf{.300}}{\mathbf{-}}{\mathbit{k}}{\mathbit{g}}$ rubber ball is dropped from a height of 1.75 m, it bounces off a concrete floor and rebounds to a height of ${\mathbf{1}}{\mathbf{.50}}{\mathbf{}}{\mathbf{}}{\mathbit{m}}$. (a) Determine the magnitude and direction of the impulse delivered to the ball by the floor. (b) Estimate the time the ball is in contact with the floor and use this estimate to calculate the average force the floor exerts on the ball.

(a) The magnitude and direction of the impulse delivered to the ball by the floor is$3.38\stackrel{^}{j}\text{\hspace{0.17em}kg}\cdot \text{m/s}$

.

(b) The average force the floor exerts on the ball is $7×{10}^{2}\stackrel{^}{j}\text{\hspace{0.17em}}\text{N}$.

See the step by step solution

## Step 1: Concept

The impulse imparted to a particle by a net forcecan be calculated as the time integral of the force

$\stackrel{\to }{I}=\underset{{t}_{i}}{\overset{{t}_{f}}{\int }}\stackrel{\to }{F}dt$

where

$\stackrel{\to }{I}=$ Impulse delivered to object

$\stackrel{\to }{F}=$ Average friction force

$\Delta t=$ Time interval in seconds.

The impulse-momentum theorem can help us to find the connection between the change in the horizontal momentum of the object to the horizontal force acting on it.

$\Delta {P}_{x}={F}_{x}\Delta t$

where

$\Delta {P}_{x}=$ Change in the horizontal momentum

${F}_{x}=$ Average friction force

$\Delta t=$ Time interval in seconds.

## Step 2:Stating given data

The data given are listed below as

• Mass of the ball, $m=0.300\text{\hspace{0.17em}kg}$
• The ball dropped from a height ${h}_{1}=1.75\text{\hspace{0.17em}}\text{m}$ and rebounded to a height ${h}_{2}=1.50\text{\hspace{0.17em}m}$ .

## Step 3: Magnitude and direction of the impulse delivered to the ball by the floor

Part (a):

The impulse the floor exerts on the ball is equal to the change in momentum of the ball.

The magnitude and direction of the impulse delivered to the ball by the floor can be calculated as

$\begin{array}{c}\Delta \stackrel{\to }{P}=m\left({\stackrel{\to }{v}}_{f}-{\stackrel{\to }{v}}_{i}\right)\\ =m\left({v}_{f}-{v}_{i}\right)\stackrel{^}{j}\end{array}$

We know if the initial speed is zero, then the relationship between the final velocity and the height is

$v=\sqrt{2gh}$

Then the final and initial velocity can be calculated as

${v}_{i}=\sqrt{2g{h}_{1}}=\sqrt{2×9.8×1.75}=5.856\approx 5.86\text{\hspace{0.17em}m/s}\phantom{\rule{0ex}{0ex}}{v}_{f}=\sqrt{2g{h}_{2}}=\sqrt{2×9.8×1.50}=5.422\approx 5.42\text{\hspace{0.17em}m/s}$

Substitute the values in the above expression, we get

$\begin{array}{c}\Delta \stackrel{\to }{P}=\left(0.300\text{\hspace{0.17em}kg}\right)\left[\left(5.42\text{\hspace{0.17em}m/s}\right)-\left(-5.86\text{\hspace{0.17em}m/s}\right)\right]\stackrel{^}{j}\\ =3.38\stackrel{^}{j}\text{\hspace{0.17em}kg}\cdot \text{m/s}\end{array}$

Thus, the magnitude and direction of the impulse delivered to the ball by the floor is $3.38\stackrel{^}{j}\text{\hspace{0.17em}kg}\cdot \text{m/s}$

## Step 4: Estimation of the time and finding the average force

Part (b):

Estimating the contact time interval to be$0.05\text{\hspace{0.17em}}s$ .

So from the impulse-momentum theorem, we can calculate the average force as

$\stackrel{\to }{F}=\frac{\Delta \stackrel{\to }{P}}{\Delta t}$

Substitute the values in the above expression, we get

$\begin{array}{c}\stackrel{\to }{F}=\frac{3.38\text{\hspace{0.17em}}\text{kg}\cdot \text{m/s}}{0.05\text{\hspace{0.17em}s}}\\ \stackrel{\to }{F}=7×{10}^{2}\stackrel{^}{j}\text{\hspace{0.17em}}\text{N}\end{array}$

Thus, the average force the floor exerts on the ball is$7×{10}^{2}\stackrel{^}{j}\text{\hspace{0.17em}}\text{N}$ . ### Want to see more solutions like these? 