Subfield: Mechanics

$g=9.8 \mathrm{~N} / \mathrm{kg}$



$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$



$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$



$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$



$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$



$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$



$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$



$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$



$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$



$$

\begin{aligned}

& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

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

& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

& e=1.602 \times 10^{-19} \mathrm{C} \\

& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

& \ln (1+x) \approx x \text { for }|x| \ll 1 \\

& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{aligned}

$$

$$

\begin{array}{lrl}

\text { velocity parameter } & \beta & =\frac{v}{c} \\

\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\

\text { relativistic momentum } & p & =\gamma \beta m c \\

\text { relativistic energy } & E & =\gamma m c^{2} \\

\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}

\end{array}

$$

Extra Supplementary Reading Materials:

Suppose a domino stands upright on a table. It has height $h$, thickness $t$, width $w$ (as shown below), and mass $m$. The domino is free to rotate about its edges, but will not slide across the table.

<img_4527>
Context question:
a. Suppose we give the domino a sharp, horizontal impulsive push with total momentum $p$.

i. At what height $H$ above the table is the impulse $p$ required to topple the domino smallest?



ii. What is the minimum value of $p$ to topple the domino?
Context answer:
\boxed{$H=h$ , $p_{\min }=\frac{1}{\sqrt{3}} \frac{m}{h} \sqrt{g\left(\sqrt{t^{2}+h^{2}}-h\right)\left(h^{2}+t^{2}\right)}$}


Question: b. Next, imagine a long row of dominoes with equal spacing $l$ between the nearest sides of any pair of adjacent dominos, as shown above. When a domino topples, it collides with the next domino in the row. Imagine this collision to be completely inelastic. What fraction of the total kinetic energy is lost in the collision of the first domino with the second domino?

Solution:
Right after the collision, the dominoes touch at a height $\sqrt{h^{2}-l^{2}}$ above the table. The second domino is vertical, while the first is rotated so that the angle between its leading edge and the table is $\theta=\arccos (l / h)$. Let the dominoes' angular velocities be $\omega_{1}$ and $\omega_{2}$ respectively. After the dominoes collide, they stick together. If we take the two parts of the dominoes that are in contact, they must have the same horizontal velocity component in order for the dominoes to stay in contact. For the first domino, this velocity is $\omega_{1} h \sin \theta=\omega_{1} \sqrt{h^{2}-l^{2}}$. For the second domino it is $\omega_{2}$ times the height of the impact point, so $\omega_{2} \sqrt{h^{2}-l^{2}}$. Because these velocity components must be equal, $\omega_{1}=\omega_{2}$.



This means the dominoes have the same angular momentum as each other, measured relative to their respective rotation axes. If we look at the collision, the forces between the dominoes are purely horizontal because the dominoes' faces are frictionless, so they only exert normal forces. The second domino is vertical, so all its normal forces are purely horizontal. These horizontal normal forces exchange angular momentum between the two dominoes. They have the same moment arm (i.e. the height of the collision), so the amount of angular momentum transferred out of the first domino is equal to the amount gained by the second domino, both measured relative to the dominoes' respective rotation axes. Because the dominoes are identical and have the same angular velocity, they have the same angular momentum. This means each domino has half as much angular momentum about its rotation axis as the first domino had just before the collision. (Note that forces from the table cannot change the angular momentum of the dominoes about their respective rotational axes because such forces have zero moment arm, so only the inter-domino forces need to be examined here.)



Kinetic energy scales with the square of angular momentum, so each domino has a quarter as much kinetic energy after the collision as the first domino had before it. That means the total kinetic energy after the collision is half what it was before the collision. The fraction of kinetic energy lost is one half.

Final answer: 50 percent