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 frictionless track contains a loop of radius $R=0.5 \mathrm{~m}$. Situated on top of the track lies a small ball of mass $m=2 \mathrm{~kg}$ at a height $h$. It is then dropped and collides with another ball (of negligible size) of mass $M=5 \mathrm{~kg}$.

<img_4268>

Let $h$ be the minimum height that $m$ was dropped such that $M$ would be able to move all the way around the loop. The coefficient of restitution for this collision is given as $e=\frac{1}{2}$.

Now consider a different scenario. Assume that the balls can now collide perfectly inelastically, which means that they stick to each other instantaneously after collision for the rest of the motion. If $m$ was dropped from a height $3 R$, find the minimum value of $\frac{m}{M}$ such that the combined mass can fully move all the way around the loop. Let this minimum value be $k$. Compute $\alpha=\frac{k^{2}}{h^{2}}$. (Note that this question is only asking for $\alpha$ but you need to find $h$ to find $\alpha$ ). Assume the balls are point masses (neglect rotational effects).

Solution:
The velocity of $m$ when it gets to the bottom of the track will be given by
$$
v_{b}=\sqrt{2 g h} .
$$

Claim: The velocity of $M$ after collision will be given by $v_{M}=\frac{(1+e) m}{m+M} \sqrt{2 g h}$.

Proof: Conservation of momentum before and after the collision is expressed by:

$$
m v_{b}=m v_{m}+M v_{M}
$$

By coefficient of restitution,

$$
v_{M}-v_{m}=e v_{b}
$$

These equations may be solved directly to find $v_{m}, v_{M}$ to give

$$
v_{M}=\frac{(1+e) m}{m+M} v_{b}
$$

Next, when the objects get to the top of the loop, conservation of energy gives the speed of $M$ when it gets to the top as

$$
\frac{1}{2} v_{M}^{2}=\frac{1}{2} v_{t}^{2}+2 g R .
$$

At the top of the loop, $M$ must at least have an acceleration $g$ to maintain circular motion, and thus

$$
\frac{v_{t}^{2}}{R}=g \Longrightarrow v_{t}=\sqrt{g R}
$$



Substituting these results into our conservation equation gives us

$$
\begin{aligned}
\frac{(1+e)^{2} m^{2}}{(m+M)^{2}}(\sqrt{2 g h})^{2} & =g R+4 g R \\
\frac{2(1+e)^{2} m^{2} g h}{(m+M)^{2}} & =5 g R \\
h & =\frac{5(m+M)^{2} R}{2(1+e)^{2} m^{2}} \\
& =6.805 \mathrm{~m}
\end{aligned}
$$

In the second scenario, conservation of momentum before and after the collision gives:

$$
m v_{b}=(M+m) v_{f} \Longrightarrow v_{f}=\frac{m v_{b}}{M+m}
$$

The same conservation of energy formula as in the first scenario yields

$$
\begin{aligned}
\frac{m^{2}}{(M+m)^{2}}(\sqrt{2 g h})^{2} & =5 g R \\
\frac{2 m^{2} g h}{(M+m)^{2}} & =5 g R \\
\frac{6 m^{2} g R}{(M+m)^{2}} & =5 g R \\
\left(\frac{k}{k+1}\right)^{2} & =\frac{5}{6} \\
\frac{k}{k+1} & =\sqrt{\frac{5}{6}} \\
k & =\frac{\sqrt{5}}{\sqrt{6}-\sqrt{5}} \\
& =10.477
\end{aligned}
$$

Thus,

$$
\alpha=\frac{k^{2}}{h^{2}}=2.37 \mathrm{~m}^{2}
$$

Final answer: $2.37$ $\mathrm{~m}^{2}$