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: A small toy car rolls down three ramps with the same height and horizontal length, but different shapes, starting from rest. The car stays in contact with the ramp at all times and no energy is lost. Order the ramps from the fastest to slowest time it takes for the toy car to drop the full $1 \mathrm{~m}$. For example, if ramp 1 is the fastest and ramp 3 is the slowest, then enter 123 as your answer choice.

Figure 1

<img_4323>

Figure 2

<img_4295>

Figure 3

<img_4275>

Solution:
One can solve this by finding the time it takes for each ramp. For ramp 1:
$$
\begin{aligned}
\sqrt{2} & =\frac{1}{2} g \sin \left(45^{\circ}\right) t^{2} \\
\Longrightarrow t & =0.639 \mathrm{~s}
\end{aligned}
$$

For ramps 2, let the length of the dashed region be $x$. Then:

$$
x+x / \tan \left(30^{\circ}\right)=1 \Longrightarrow x=0.366 \mathrm{~m}
\tag{1}
$$

Due to symmetry, both the steep and shallow regions of both ramps 2 and 3 have a length of $x / \cos \left(60^{\circ}\right)=0.732 \mathrm{~m}$. This results in a time for ramp 2 as:

$$
\begin{aligned}
& 0.732=\frac{1}{2} g \sin \left(60^{\circ}\right) t_{1}^{2} \Longrightarrow t_{1}=0.415 \mathrm{~s} \\
& 0.732=\sqrt{2 g(1-x)} t_{2}+\frac{1}{2} g \sin \left(30^{\circ}\right) t_{2}^{2} \Longrightarrow t_{2}=0.184 \mathrm{~s}
\end{aligned}
$$

for a total time of $t_{1}+t_{2}=0.599 \mathrm{~s}$. For ramp 3 ,

$$
\begin{aligned}
& 0.732=\frac{1}{2} g \sin \left(30^{\circ}\right) t_{1}^{2} \Longrightarrow t_{1}=0.546 \mathrm{~s} \\
& 0.732=\sqrt{2 g(1-x)} t_{2}+\frac{1}{2} g \sin \left(60^{\circ}\right) t_{2}^{2} \Longrightarrow t_{2}=0.206 \mathrm{~s}
\end{aligned}
$$

which gives a total time of $t_{1}+t_{2}=0.752 \mathrm{~s}$. From fastest to slowest, the answer becomes 213 . Note that this answer is easily guessable via intuition.

Final answer: 213