Subfield: Mechanics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Poncho is a very good player of the legendary carnival game known as Pico-Pico. Its setup consists of a steel ball, represented by a point mass, of negligible radius and a frictionless vertical track. The goal of Pico-Pico is to flick the ball from the beginning of the track (point $A$ ) such that it is able to traverse through the track while never leaving the track, successfully reaching the end (point $B)$. The most famous track design is one of parabolic shape; specifically, the giant track is of the shape $h(x)=5-2 x^{2}$ in meters. The starting and ending points of the tracks are where the two points where the track intersects $y=0$. If $\left(v_{a}, v_{b}\right]$ is the range of the ball's initial velocity $v_{0}$ that satisfies the winning condition of Pico-Pico, help Poncho find $v_{b}-v_{a}$. This part is depicted below:
<img_4262>

Solution:
Using conservation of energy, the minimum initial velocity of the ball needed to pass the top of the track is $v_{a}=\sqrt{2 g h}=9.9045 \frac{\mathrm{m}}{\mathrm{s}}$. To find $v_{b}$, the centripetal force at all points on the track must be determined given the initial velocity.
$$

F_{c} =\frac{m v^{2}}{R}
\tag{11}
$$
$$
=\frac{m\left(v_{b}^{2}-2 g h\right)}{\frac{\left|1+\left(\frac{d}{d x} h(x)\right)^{2}\right|}{\frac{d^{2}}{d x^{2}} h(x)}}
\tag{12}
$$
$$
=\frac{m\left(v_{b}^{2}-2 g h\right)}{\frac{\left|1+16 x^{2}\right|^{\frac{3}{2}}}{4}}
\tag{13}
$$

For the boundary condition, the ball leaves if the normal force from the track on the ball $N=m g \cos \theta-F_{c}$ becomes 0.

$$
\begin{aligned}
m g \cos \theta-F_{c} & =0 \\
m g \cos \arctan (-4 x) & =\frac{4 m\left(v_{b}^{2}-2 g h\right)}{\left|1+16 x^{2}\right|^{\frac{3}{2}}} \\
\frac{g}{\left|1+16 x^{2}\right|^{\frac{1}{2}}} & =\frac{4\left(v_{b}^{2}-2 g h\right)}{\left|1+16 x^{2}\right|^{\frac{3}{2}}} \\
g & =\frac{4\left(v_{b}^{2}-2 g h\right)}{1+16 x^{2}} \\
v_{b \max } & =\sqrt{\frac{g+16 g x^{2}}{4}+2 g h}
\end{aligned}
$$

From the derivation, $v_{b \max }$ is the lowest at $x=0$. Thus,

$$
\begin{aligned}
v_{b \max } & =\sqrt{\frac{g}{4}+2 g h} \\
& =10.0276 \frac{\mathrm{m}}{\mathrm{s}}
\end{aligned}
$$

which is our desired $v_{b}$. The final answer, $v_{b}-v_{a}$, can be calculated.

Final answer: 0.1231 m/s