Home Page Textbooks Physics: Principles and Problems Page 301: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

All Solutions

Page 301: Section Review

Exercise 22

Solution 1

Solution 2

Step 1

1 of 2

We need to determine whether the Earth is a closed, isolated system or not.

Step 2

2 of 2

Depending on how we set up the limits of our system and what what types of energy we are considering in each problem we can act as if Earth is an isolated, closed system. But taking everything into consideration, the Earth is not a closed, isolated system because there is an exchange of both mass and energy with its environment such as other planets, stars, comets, space itself…

Step 1

1 of 2

No, because external forces from the surrounding celestial bodies, including the Sun and the other planets, are acting on the Earth

Result

2 of 2

No, because external forces from the surrounding celestial bodies, including the Sun and the other planets, are acting on the Earth

Exercise 23

Step 1

1 of 3

noindent
(a) At the highest point, the potential energy of the child is at the maximum while the kinetic energy is at the minimum.\
(b) At the lowest point, the kinetic energy of the child is at the maximum while the potential energy is at the minimum.\

Step 2

2 of 3

Result

3 of 3

Exercise 24

Step 1

1 of 2

The kinetic energy will not be conserved because the collision between the chewing gum and rubber ball is inelastic. The kinetic energy loss is converted to thermal energy in the form of vibrational energy of the atoms.

Result

2 of 2

Exercise 25

Solution 1

Solution 2

Step 1

1 of 2

The racquet and the ball are designed that way to dissipate the kinetic energy of the ball as it bounce off the racquet allowing the players to have more control of the trajectory of the ball.

Baseball also have this kind of ball-paddle pair interaction.The ball used in baseball is soft while the bat used to hit this ball is significantly harder.

Result

2 of 2

The racquet and the ball are designed that way to dissipate the kinetic energy of the ball as it bounce off the racquet allowing the players to have more control of the trajectory of the ball.

Baseball also have this kind of ball-paddle pair interaction.The ball used in baseball is soft while the bat used to hit this ball is significantly harder.

Step 1

1 of 2

The balls, paddle, and racquet are designed to match so that you can produce a maximum amount of kinetic energy once the paddle or racquet hit the ball.

Step 2

2 of 2

Softer balls recieve energy with less less from the paddle or racket.

Exercise 26

Solution 1

Solution 2

Step 1

1 of 5

In this problem we need to consider a rubber ball being dropped from a height of $h_0=8,,rm{m}$. It bounces off the floor, and each time it bounces, it loses $frac{1}{5}$ of its energy. We need to determine how many times it can bounce before being unable to bounce to a height of $h=4,,rm{m}$

Step 2

2 of 5

If by each bounce it loses $frac{1}{5}$ of its energy, it means that it retains other $frac{4}{5}$. Energy that we are considering here is potential energy which can be expressed as:

$$
E_p=mgh
$$

If the mass $m$ and gravitational constat stay equal, the only change that happens is in height. That means that we can express connection between bounces by next equation:

$$
E_0=E_1cdot (frac{4}{5})^1
$$

Instead of energies, we can use height:

$$
h_0=h_1cdot (frac{4}{5})^1
$$
.

Step 3

3 of 5

And for more consequent bounces we can write the next equation:

$$
h_n=h_0 cdot (frac{4}{5})^n
$$

Where $n$ is a number of bounces. In our problem we need to determine $n$ that will provide us with $h_n>4,,rm{m}$

Step 4

4 of 5

When we insert values we get:

$$
4=8cdot (frac{4}{5})^n
$$

Dividing both sideds by 4 we get:

$$
0.5=(0.8)^n
$$

Using logarithm we can get:

$$
log_{0.8}(0.5)=n
$$

From which we can get:

$$
n=3.1
$$

This means that ball will make $3.1$ bounces before reaching height of $4$ m, since we can only count full bounces, this means that the ball can make $textbf{three bounces}$ before bouncing to a height of less that $4$ m.

Result

5 of 5

$$
n=3
$$

Step 1

1 of 2

noindent
textbf{Given:}\
$h_o = 8.0 text{m}$ \

noindent
textbf{Calculation:}\
noindent
Since the potential energy is directly proportional to the height (as shown in the equation below), change in potential energy of the ball will also cause change in its height with the same amount.

$$
PE = mgh
$$
noindent
On the first bounce, the new height is calculated by

$$
h_1 = h_0left( frac{4}{5} right)
$$

$$
h_1 = (8.0)cdotleft( frac{4}{5} right)
$$

$$
h_1 = 6.4 text{m}
$$

noindent
On the second bounce, the new height is calculated by

$$
h_2 = h_1left( frac{4}{5} right)
$$

$$
h_2 = (6.4)cdotleft( frac{4}{5} right)
$$

$$
h_2 = 5.12 text{m}
$$

noindent
On the third bounce, the new height is calculated by

$$
h_3 = h_2left( frac{4}{5} right)
$$

$$
h_3 = (5.12)cdotleft( frac{4}{5} right)
$$

$$
h_3 = 4.10 text{m}
$$

noindent
It took $boxed{3 text{bounces}}$ until the ball bounces back up to a height of $4 text{m}$.

Result

2 of 2

$3 text{bounces}$

Exercise 27

Solution 1

Solution 2

Step 1

1 of 4

Child weighing $m=36,,rm{kg}$ starts its slide from the height of $h=2.5,,rm{m}$, and reaches velocity of $v=3,,rm{m/s}$ at the bottom of the slide.We need to determine the energy that was lost sliding.

Step 2

2 of 4

Energy that was lost can be calculated throught the difference in the initial and the final energy. Initially the child is stationary so the only energy is the potential energy of that child at a height of $h=2.5,,rm{m}$. At the bottom there is no more potential energy because all of it has either been converted to the kinetic energy or has been lost.

$$
Delta E=E_p-E_l
$$

Step 3

3 of 4

Inserting equations for potential and kinetic energy into the previous equation we can get:

$$
Delta E=mgh-0.5cdot mv^2
$$

Inserting values we get:

$$
Delta E=36cdot 9.8cdot 2.5 – 0.5cdot 36cdot 3^2
$$

Finally we get:

$$
boxed{Delta E=720,,rm{J}}
$$

Result

4 of 4

$$
Delta E=720,,rm{J}
$$

Step 1

1 of 2

The initial energy of the child is:

$E_i = PE = m g h = 36.0 * 9.80 * 2.5 = 882 J$

The final energy of the child is:

$E_f = KE = (1/2) m v^2 = (1/2) * 36.0 * (3.0)^2 = 162 J$

The amount of loosed energy:

$$
E_f – E_i = 162 – 882 = -720 J
$$

Result

2 of 2

$$
-720 J
$$

Exercise 28

Solution 1

Solution 2

Step 1

1 of 2

We need to consider the falling ball and the scenario in which the ball in free fall falls for only half the amount of time it take her to fall completely and have to determine whether the amount of potential energy of the ball at that point will be more than, equal to or less than a half of the ball’s total energy.

Step 2

2 of 2

The longer the ball falls, the more potential energy converts to the kinetic energy and therefore increases velocity. This means that in free fall the ball is constantly accelerating.

This further means that it takes the ball longer to cover the first half of the distance than it does to cover a second half since it has higher velocity in the second half. This means that, at a half of time it takes the ball to fall completely, the ball $textbf{hasn’t yet reached a halfway point in distance}$.

This further means that less than a half of the potential energy has been converted to the kinetic, which further means that $textbf{more than a half}$ of the balls total energy is still in the form of potential energy.

Step 1

1 of 2

noindent
When the ball has fallen for half the amount of time it takes to fall, less than half of its energy will be potential energy. This can be better visualized using the equation below.

$$
d = frac{1}{2}gt^2
$$

The distance travelled by the ball as it drops is directly proportional to $t^2$. Thus, if the ball is already travelling for half of its travel time, it already covers more than half of the distance from the drop-off point to the ground making its potential energy significantly lower.

Result

2 of 2

When the ball has fallen for half the amount of time it takes to fall, less than half of its energy will be potential energy.

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Page 301: Section Review

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