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{$H=h$ , $p_{\min }=\frac{1}{\sqrt{3}} \frac{m}{h} \sqrt{g\left(\sqrt{t^{2}+h^{2}}-h\right)\left(h^{2}+t^{2}\right)}$}


Context 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?
Context answer:
\boxed{50}


Question: c. After the collision, the dominoes rotate in such a way so that they always remain in contact. Assume that there is no friction between the dominoes and the first domino was given the smallest possible push such that it toppled. What is the minimum $l$ such that the second domino will topple?



You may work to lowest nontrivial order in the angles through which the dominoes have rotated. Equivalently, you may approximate $t, l \ll h$.

Solution:
As in part (a), we will find the highest potential energy of the system and make sure the initial kinetic energy is high enough to get the system to that point.



At any time after the collision, let us call the angle that the first domino has rotated past its point of highest potential energy $\alpha$, and the angle that the second domino has rotated past its point of highest potential energy $\beta$. Also, let's call the angle a domino rotates from its standing position up to its point of highest potential energy $\phi$, so the dominoes have rotated $\phi+\alpha$ and $\phi+\beta$ respectively.



The dominoes need to be in contact. The top right corner of the first domino has moved horizontally a distance $h \sin (\phi+\alpha)$ which we will approximate as $h(\phi+\alpha)$ using the small angle approximation. The top left corner of the second domino moves horizontally forward by $h \sin (\phi+\beta)+t(1-\cos (\phi+\beta))$. We ignore the cosine term as second order in the rotation angle and approximate this as $h(\phi+\beta)$.



The $y$-coordinate of the upper right corner of the first domino is $h(1-\cos (\phi+\alpha)) \approx h$. The $y$-coordinate of the upper left corner of the second domino is $h(1-\cos (\phi+\beta))+t \sin (\phi+\beta) \approx$ $h+t(\phi+\beta)$ There is a first-order difference in $y$-coordinates of the two corners, but this means the difference in $x$ coordinate between the top left corner of the second domino and the top right corner of the first domino is second-order in the rotation angles. We conclude that to first order



$$

h(\phi+\alpha)=l+h(\phi+\beta) \quad \Rightarrow \quad \alpha=\frac{l}{h}+\beta

$$



because this condition puts the top right corner of the first domino at the same position as the top left of the second domino.



The potential energy of the first domino, setting zero potential energy to be when the domino is upright, is $U_{1}=\Delta U\left(1-(\alpha / \phi)^{2}\right)$ to second order in $\alpha$, and for the second domino, $U_{2}=$ $\Delta U\left(1-(\beta / \phi)^{2}\right)$. These figures come from fitting a quadratic whose peak is when the center of mass is above the rotation point and which is zero when the domino is upright. The total potential energy is $U_{1}+U_{2}$, and using the relation between $\alpha$ and $\beta$, it is minimized for



$$

\beta_{\max }=-\frac{l}{2 h}

$$



In other words, the second domino is as far away from rotating to the top of arc as the first domino has rotated past the top of its arc. This gives a maximum potential energy



$$

U_{\max }=2 \Delta U\left(1-\frac{l^{2}}{4 t^{2}}\right)

$$



where we have used the approximation $\phi \approx t / h$.


At impact, $U_{\text {impact }} \approx \Delta U\left(-l^{2} / t^{2}+2 l / t\right)$. Before impact, the kinetic energy is $\Delta U-U_{\text {impact }}$ because the maximum potential energy before impact was $\Delta U$, and the kinetic energy was zero there. The kinetic energy just after the collision is then $\left(\Delta U-U_{\text {impact }}\right) / 2$. Setting this equal to the potential energy gain as the two dominoes rotate to their highest potential energy $U_{\max }$,



$$

\frac{1}{2}\left(\Delta U-U_{\text {impact }}\right)=U_{\max }-U_{\text {impact }}

$$



or



$$

\frac{1}{2} \Delta U=U_{\max }-\frac{1}{2} U_{\text {impact }}

$$



Plugging in the earlier expressions for all these gives



$$

\frac{1}{2} \Delta U=2 \Delta U\left(1-\frac{l^{2}}{4 t^{2}}\right)-\Delta U\left(\frac{l}{t}-\frac{l^{2}}{2 t^{2}}\right)

$$



and solving this yields



$$

l=\frac{3}{2} t

$$



to first order in $t$.

Final answer: $l=\frac{3}{2} t$