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Q14 P

Expert-verifiedFound in: Page 284

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Review: After a ${\mathbf{0}}{\mathbf{.300}}{\mathbf{-}}{\mathit{k}}{\mathit{g}}$ rubber ball is dropped from a height of 1.75 m, it bounces off a concrete floor and rebounds to a height of ${\mathbf{1}}{\mathbf{.50}}{\mathbf{}}{\mathbf{}}{\mathit{m}}$. **

**(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$3.38\widehat{j}\text{\hspace{0.17em}kg}\cdot \text{m/s}$

.

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

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

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

where

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

$\overrightarrow{F}=$ 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.

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 ${h}_{2}=1.50\text{\hspace{0.17em}m}$ .

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

$\begin{array}{c}\Delta \overrightarrow{P}=m({\overrightarrow{v}}_{f}-{\overrightarrow{v}}_{i})\\ =m({v}_{f}-{v}_{i})\widehat{j}\end{array}$

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

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

Substitute the values in the above expression, we get

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

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

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

$\overrightarrow{F}=\frac{\Delta \overrightarrow{P}}{\Delta t}$

Substitute the values in the above expression, we get

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

Thus, the average force the floor exerts on the ball is$7\times {10}^{2}\widehat{j}\text{\hspace{0.17em}}\text{N}$ .

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