di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Mechanics

LIGO-GW150914

In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.

Part A: Newtonian (conservative) orbits
Context question:
A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,

$$
M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
\tag{!}
$$

The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as

$$
\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
\tag{2}
$$

where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.
Context answer:
\boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$}


Context question:
A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as:

$$
E=A(\mu, \Omega, L)-G \frac{M \mu}{L},
\tag{3}
$$

where

$$
\mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2}
\tag{4}
$$

are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$.
Context answer:
\boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$}


Question: A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$.

Solution:
Energy (3) of the question sheet can be interpreted as describing a system of a mass $\mu$ in a circular orbit with angular velocity $\Omega$, radius $L$, around a mass $M$ (at rest). Equating the gravitational acceleration to the centripetal acceleration:

$$
G \frac{M}{L^{2}}=\Omega^{2} L
\tag{8}
$$

This is indeed Kepler's third law (for circular orbits). Then, from 7,

$$
E=-\frac{1}{2} G \frac{M \mu}{L}
\tag{9}
$$

Final answer: $\beta=-\frac{1}{2}$
Subfield: Mechanics


Question: B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as:

$$
Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t .
\tag{8}
$$

Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$.

Solution:
Some simple trigonometry for the $x, y$ motion of the masses (in an appropriate Cartesian system) yields:

$$
\left(x_{1}, y_{1}\right)=r_{1}(\cos (\Omega t), \sin (\Omega t)), \quad\left(x_{2}, y_{2}\right)=-r_{2}(\cos (\Omega t), \sin (\Omega t))
\tag{10}
$$

Then,

$$
Q_{i j}=\frac{M_{1} r_{1}^{2}+M_{2} r_{2}^{2}}{2}\left(\begin{array}{ccc}
\frac{4}{3} \cos ^{2}(\Omega t)-\frac{2}{3} \sin ^{2}(\Omega t) & 2 \sin (\Omega t) \cos (\Omega t) & 0 \\
2 \sin (\Omega t) \cos (\Omega t) & \frac{4}{3} \sin ^{2}(\Omega t)-\frac{2}{3} \cos ^{2}(\Omega t) & 0 \\
0 & 0 & -\frac{2}{3}
\end{array}\right)
\tag{11}
$$

or, using some simple trigonometry and (6),

$$
Q_{i j}=\frac{\mu L^{2}}{2}\left(\begin{array}{ccc}
\frac{1}{3}+\cos 2 \Omega t & \sin 2 \Omega t & 0 \\
\sin 2 \Omega t & \frac{1}{3}-\cos 2 \Omega t & 0 \\
0 & 0 & -\frac{2}{3}
\end{array}\right)
\tag{12}
$$

Final answer: $$
k=2 \Omega, \quad a_{1}=a_{2}=\frac{1}{3}, a_{3}=-\frac{2}{3}, \quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \stackrel{\text { otherwise }}{=} 0
$$
Subfield: Mechanics


Question: B.6 Estimate the minimal orbital separation between the two objects at $t_{\overline{\mathrm{CD}}}$. Hence estimate a maximum size for each object, $R_{\max }$. Obtain $R_{\odot} / R_{\max }$ to compare this size with the radius of our Sun, $R_{\odot} \simeq 7 \times 10^{5} \mathrm{~km}$. Estimate also their orbital linear velocity at the same instant, $v_{\text {col }}$, comparing it with the speed of light, $v_{\text {col }} / c$.

Solution:
From (8), Kepler's law states that $L=\left(G M / \Omega^{2}\right)^{1 / 3}$. The second pair of points highlighted in the plot correspond to the cycle prior to merger. Thus, we use (19) to obtain the orbital angular velocity at $t_{\overline{\mathrm{CD}}}$ :

$$
\Omega_{t_{\overline{\mathrm{CD}}}} \sim 2.6 \times 10^{2} \mathrm{rad} / \mathrm{s}
\tag{26}
$$

Then, using the total mass (25) we find

$$
L \sim 5 \times 10^{2} \mathrm{~km}
\tag{27}
$$

Thus, these objects have a maximum radius of $R_{\max } \sim 250 \mathrm{~km}$. Hence they have over 30 times more mass and,

$$
\frac{R_{\odot}}{R_{\max }} \sim 3 \times 10^{3}
\tag{28}
$$

they are 3000 times smaller than the Sun and!

Their linear velocity is

$$
v_{\mathrm{col}}=\frac{L}{2} \Omega \simeq 7 \times 10^{4} \mathrm{~km} / \mathrm{s}
\tag{29}
$$

They are moving at over $20 \%$ of the velocity of light!

Final answer: $$
L_{\text {collision }} \sim 5 \times 10^{2} \mathrm{~km}, \quad \frac{R_{\odot}}{R_{\max }} \sim 3 \times 10^{3}, \quad \frac{v_{\text {col }}}{c} \sim 0.2
$$
Subfield: Modern Physics

Where is the neutrino?

When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called "jet". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting.

Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$;

Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$

Part A.ATLAS Detector physics

Question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.

Solution:
The magnetic force is the centripetal force:

$$
m \frac{v^{2}}{r}=e v B \Rightarrow r=\frac{m v}{e B}
$$

First express the velocity in terms of the kinetic energy,

$$
K=\frac{1}{2} m v^{2} \Rightarrow v=\sqrt{\frac{2 K}{m}},
$$

and then insert it in the expression above for the radius to get

Final answer: $r=\frac{\sqrt{2 K m}}{e B}$
Subfield: Modern Physics

Where is the neutrino?

When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called "jet". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting.

Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$;

Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$

Part A.ATLAS Detector physics
Context question:
A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.
Context answer:
\boxed{$r=\frac{\sqrt{2 K m}}{e B}$}


Extra Supplementary Reading Materials:

Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered.

Question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$.

Solution:
The radius of the circular motion of a charged particle in the presence of a uniform magnetic field is given by,

$$
r=\frac{m v}{e B}
$$

This formula is valid in the relativistic scenario if the mass correction, $m \rightarrow \gamma m$ is included:

$$
r=\frac{\gamma m v}{e B}=\frac{p}{e B} \Rightarrow p=r e B
$$

Note that the radius of the circular motion is half the radius of the inner part of the detector. One obtains $\left[1 \mathrm{MeV} / c=5.34 \times 10^{-22} \mathrm{~m} \mathrm{~kg} \mathrm{~s}^{-1}\right.$ ]

Final answer: $p=330$ $ \mathrm{MeV} / c$
Subfield: Thermodynamics

Physics of Live Systems

Data: Normal atmospheric pressure, $P_{0}=1.013 \times 10^{5} \mathrm{~Pa}=760 \mathrm{mmHg}$

Part A.The physics of blood flow

In this part you will analyse two simplified models of blood flow in vessels.

Blood vessels are approximately cylindrical in shape, and it is known that for a steady, non turbulent flow of an incompressible fluid in a rigid cylinder, the difference in pressure of the fluid at the two ends of the cylinder is given by

$$
\Delta P=\frac{8 \ell \eta}{\pi r^{4}} Q
\tag{1}
$$

where $\ell$ and $r$ are the length and radius of the cylinder, $\eta$ is the fluid viscosity and $Q$ is the volumetric flow rate, i.e. the fluid volume that passes the cylinder cross section per unit time. This expression is often able to provide the correct order of magnitude for the pressure difference in a vessel, even without taking into account the pulsatile flow, the vessel's compressibility and irregular shape, and the fact that blood is not a simple fluid but a mixture of cells and plasmA.Moreover, this expression has the same form as Ohm's law, with the volumetric flow rate being interpreted as a current, the difference in pressure as a voltage, and the factor $R=\frac{8 \ell \eta}{\pi r^{4}}$ as a resistance.

Consider for example the symmetrical network of arterioles (small arteries) depicted in Figure 1 that delivers blood to the capillary bed of a tissue. In this network, at each bifurcation a vessel is divided in two identical vessels. However, the vessels of higher levels are thinner and shorter: consider that the radii and lengths of vessels in two consecutive levels, $i$ and $i+1$, are related by $r_{i+1}=r_{i} / 2^{1 / 3}$ and by $\ell_{i+1}=\ell_{i} / 2^{1 / 3}$.

<img_4505>

Level $\quad \begin{array}{lllllll}0 & 1 & 2 & 3 & \cdots & N-1\end{array}$

Figure 1. Network of arterioles.

Question: A.1 Obtain an expression for the volumetric flow rate, $Q_{i}$, in a vessel at any level $i$, as a function of the total number of levels $N$, of the viscosity $\eta$, of the radius $r_{0}$ and length $\ell_{0}$ of the first vessel, and of the difference $\Delta P=P_{0}-P_{\text {cap }}$ between the pressure at the arteriole at level $0, P_{0}$, and the pressure at the capillary bed, $P_{\text {cap }}$.

Solution:
Since the vessel network is symmetrical, the flow in a vessel of level $i+1$ is half the flow in a vessel of level $i$.

In this way, we can sum the pressure differences in all levels:

$$
\Delta P=\sum_{i=0}^{N-1} Q_{i} R_{i}=Q_{0} \sum_{i=0}^{N-1} \frac{R_{i}}{2^{i}}
$$

Introducing the radii dependences yields

$$
\Delta P=Q_{0} \sum_{i=0}^{N-1} \frac{8 \ell_{i} \eta}{2^{i} \pi r_{i}^{4}}=Q_{0} \frac{8 \ell_{0} \eta}{\pi r_{0}^{4}} \sum_{i=0}^{N-1} \frac{2^{4 i / 3}}{2^{i} 2^{i / 3}}=Q_{0} N \frac{8 \ell_{0} \eta}{\pi r_{0}^{4}}
$$

Therefore

$$
Q_{0}=\Delta P \frac{\pi r_{0}^{4}}{8 N \ell_{0} \eta}
$$

Final answer: $Q_{i}=\Delta P \frac{\pi r_{0}^{4}}{2^{i+3} N \ell_{0} \eta}$
Subfield: Thermodynamics


Question: A.3 Obtain, in the stationary regime, the pressure amplitude at the vessel outlet, $P_{\text {out }}$, as a function of the pressure amplitude at the inlet, $P_{\text {in }}$, the equivalent resistance, $R$, inductance, $L$ and capacitance, $C$, for a flow with angular frequency $\omega$. Establish the condition between $\eta, \rho, E, h, r$ and $\ell$ so that, for low frequencies, the pressure oscillation amplitude at the outlet is smaller than that of $P_{\text {in }}$.

Solution:
The current is given by

$$
I=\frac{P_{\mathrm{in}} \mathrm{e}^{i \omega t}}{R+i \omega L+\frac{1}{i \omega C}}
$$

The pressure difference in the capacitor is

$$
P_{\text {out }} \mathrm{e}^{i(\omega t+\phi)}=\frac{P_{\text {in }} \mathrm{e}^{i \omega t}}{R+i \omega L+\frac{1}{i \omega C}} \frac{1}{i \omega C}=\frac{P_{\text {in }} \mathrm{e}^{i \omega t}}{i \omega C R-\omega^{2} L C+1} .
$$

The amplitude is

$$
P_{\text {out }}=\frac{P_{\text {in }}}{\sqrt{\left(1-\omega^{2} L C\right)^{2}+\omega^{2} C^{2} R^{2}}} .
$$

To be smaller than $P_{\text {in }}$, for $\omega \rightarrow 0$ :

$$
\left(1-\omega^{2} L C\right)^{2}+\omega^{2} C^{2} R^{2}>1 \Longleftrightarrow-2 C L+C^{2} R^{2}>0
$$

Replacing the expressions for $L, C$, and $R$ we get: $\frac{642^{2} \ell^{2}}{3 E h r^{3} \rho}>1$.

Final answer: $$
P_{\text {out }}=\frac{P_{\text {in }}}{\sqrt{\left(1-\omega^{2} L C\right)^{2}+\omega^{2} C^{2} R^{2}}} .
$$

Condition:

$$
\frac{64 \eta^{2} \ell^{2}}{3 E h r^{3} \rho}>1
$$
Subfield: Mechanics

## Toffee Pudding



A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over.

Question: a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.

Solution:
This is when the spring force equals the maximal static friction, $k s=\mu_{0} m g$, so $s=\mu_{0} m g / k$.

Final answer: $s=\mu_{0} m g / k$
Subfield: Mechanics

## Toffee Pudding



A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over.
Context question:
a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.
Context answer:
\boxed{$s=\mu_{0} m g / k$}


Question: b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving.

Solution:
The net stretching of the spring is $s+v_{0} t-x$, leading to a rightward force $k s$. When the box is moving, it is always moving to the right, so the kinetic friction force $\mu m g$ is always in the leftward direction, which means



$$

m a=k\left(s+v_{0} t-x\right)-\mu m g

$$



which means



$$

a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g .

$$

Final answer: $a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g$
Subfield: Mechanics

## Toffee Pudding



A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over.
Context question:
a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.
Context answer:
\boxed{$s=\mu_{0} m g / k$}


Context question:
b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving.
Context answer:
\boxed{$a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g$}


Extra Supplementary Reading Materials:

The position of the box as a function of time $t$ as defined in part (b) is



$$

x(t)=\frac{v_{0}}{\omega}(\omega t-\sin \omega t)+(1-r) s(1-\cos \omega t),

$$



where $\omega=\sqrt{k / m}$ and $r=\mu / \mu_{0}$. This expression applies as long as the box is still moving, and you can use it in the parts below. Express all your answers in terms of $v_{0}, \omega, s$, and $r$.

Question: c. Find the time $t_{0}$ when the box stops for the first time.

Solution:
Taking the derivative, the velocity of the box is



$$

v=v_{0}(1-\cos \omega t)+(1-r) s \omega \sin \omega t .

$$



The box stops when this is equal to zero for the first time. There are several ways to







evaluate this condition. First, we can use half-angle identities to find



$$

0=2 v_{0} \sin ^{2} \frac{\omega t}{2}+2(1-r) s \omega \sin \frac{\omega t}{2} \cos \frac{\omega t}{2}

$$



As a result, the box stops when



$$

\tan \frac{\omega t}{2}=-\frac{(1-r) s \omega}{v_{0}}

$$



Using a basic property of the tangent function,



$$

\tan \left(\pi-\frac{\omega t}{2}\right)=\frac{(1-r) s \omega}{v_{0}}

$$



Solving for $t$, we conclude that



$$

t_{0}=\frac{2 \pi-2 \alpha}{\omega}, \quad \alpha=\tan ^{-1} \frac{(1-r) s \omega}{v_{0}}

$$



Note that we cancelled a factor of $\sin (\omega t / 2)$, which has a zero at $t=2 \pi / \omega$. However, this is a larger time than the one we just found, so it is irrelevant.



Another way to arrive at the answer is to rewrite the original condition as



$$

1=\cos \omega t-\tan \alpha \sin \omega t

$$



Squaring both sides and using some trigonometric identities gives



$$

\tan \omega t=-\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}

$$



This can then be further simplified using the tangent half-angle identity, upon which we recover the same result as above.

Final answer: $t_{0}=\frac{2 \pi-2 \alpha}{\omega}, \quad \alpha=\tan ^{-1} \frac{(1-r) s \omega}{v_{0}}$
Subfield: Mechanics


Question: d. For what values of $r$ will the spring always be at least as long as its rest length?

Solution:
The spring is stretched by $\Delta \ell=s+v_{0} t-x$. Inserting the solution for $x$, we have



$$

\Delta \ell=r s+\frac{v_{0}}{\omega} \sin \omega t+(1-r) s \cos \omega t .

$$



The most convenient way to write this is to use the sine addition formula in reverse, getting



$$

\Delta \ell=r s+\frac{v_{0}}{\omega \cos \alpha} \sin (\omega t+\alpha)

$$



The minimum stretch thus occurs when



$$

\omega t+\alpha=\frac{3 \pi}{2}

$$







Of course, we should check that this time is before the box stops; comparing with the answer to part (c) shows that it is. For the spring to always be as long as its rest length, we need the stretch at this time to be nonnegative,



$$

r s-\frac{v_{0}}{\omega \cos \alpha} \geq 0

$$



Solving the triangle, we have



$$

\cos \alpha=\frac{v_{0}}{\sqrt{v_{0}^{2}+((1-r) s \omega)^{2}}}

$$



Plugging this in and simplifying gives the answer,



$$

r \geq \frac{1}{2}\left(1+\left(\frac{v_{0}}{s \omega}\right)^{2}\right)

$$



Note that if $v_{0} / s \omega$ is too large, then it is impossible to satisfy this condition, since we need to have $r<1$.

Final answer: $r \geq \frac{1}{2}\left(1+\left(\frac{v_{0}}{s \omega}\right)^{2}\right)$
Subfield: Mechanics

## Toffee Pudding



A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over.
Context question:
a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.
Context answer:
\boxed{$s=\mu_{0} m g / k$}


Context question:
b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving.
Context answer:
\boxed{$a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g$}


Extra Supplementary Reading Materials:

The position of the box as a function of time $t$ as defined in part (b) is



$$

x(t)=\frac{v_{0}}{\omega}(\omega t-\sin \omega t)+(1-r) s(1-\cos \omega t),

$$



where $\omega=\sqrt{k / m}$ and $r=\mu / \mu_{0}$. This expression applies as long as the box is still moving, and you can use it in the parts below. Express all your answers in terms of $v_{0}, \omega, s$, and $r$.
Context question:
c. Find the time $t_{0}$ when the box stops for the first time.
Context answer:
\boxed{$t_{0}=\frac{2 \pi-2 \alpha}{\omega}, \quad \alpha=\tan ^{-1} \frac{(1-r) s \omega}{v_{0}}$}


Context question:
d. For what values of $r$ will the spring always be at least as long as its rest length?
Context answer:
$r \geq \frac{1}{2}\left(1+\left(\frac{v_{0}}{s \omega}\right)^{2}\right)$


Question: e. After the box stops, how long will it stay at rest before starting to move again?

Solution:
Using a result we found in part (d), the stretch is



$$

\Delta \ell=r s+\frac{v_{0}}{\omega \cos \alpha} \sin \left(\omega t_{0}+\alpha\right)

$$



when the box stops. Plugging in the value of $t_{0}$ found in part (c),



$$

\Delta \ell=r s+\frac{v_{0}}{\omega \cos \alpha} \sin (2 \pi-\alpha)=r s-\frac{v_{0}}{\omega} \tan \alpha=(2 r-1) s

$$



The box starts to move again when the stretch becomes $s$, so the time is



$$

\frac{s-(2 r-1) s}{v_{0}}=\frac{2(1-r) s}{v_{0}}

$$



The pattern of motion investigated in this problem is known as "stick-slip" and occurs in many practical contexts.

Final answer: $\frac{2(1-r) s}{v_{0}}$