All Solutions
Page 301: Section Review
(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.\
(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.
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.
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.
$$
E_p=mgh
$$
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:
$$
h_0=h_1cdot (frac{4}{5})^1
$$
.
$$
h_n=h_0 cdot (frac{4}{5})^n
$$
Where $n$ is a number of bounces. In our problem we need to determine $n$ that will provide us with $h_n>4,,rm{m}$
$$
4=8cdot (frac{4}{5})^n
$$
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.
n=3
$$
textbf{Given:}\
$h_o = 8.0 text{m}$ \
noindent
textbf{Calculation:}\
noindent
Since the potential energy is directly proportional to the height (as shown in the equation below), change in potential energy of the ball will also cause change in its height with the same amount.
$$
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}
$$
noindent
It took $boxed{3 text{bounces}}$ until the ball bounces back up to a height of $4 text{m}$.
$$
Delta E=E_p-E_l
$$
$$
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}}
$$
Delta E=720,,rm{J}
$$
$E_i = PE = m g h = 36.0 * 9.80 * 2.5 = 882 J$
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:
$$
E_f – E_i = 162 – 882 = -720 J
$$
-720 J
$$
This further means that it takes the ball longer to cover the first half of the distance than it does to cover a second half since it has higher velocity in the second half. This means that, at a half of time it takes the ball to fall completely, the ball $textbf{hasn’t yet reached a halfway point in distance}$.
This further means that less than a half of the potential energy has been converted to the kinetic, which further means that $textbf{more than a half}$ of the balls total energy is still in the form of potential energy.
When the ball has fallen for half the amount of time it takes to fall, less than half of its energy will be potential energy. This can be better visualized using the equation below.
$$
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.
