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

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1

\end{array}

$$

Extra Supplementary Reading Materials:

Green Revolution ${ }^{1}$



In this problem, we will investigate a simple thermodynamic model for the conversion of solar energy into wind. Consider a planet of radius $R$, and assume that it rotates so that the same side always faces the Sun. The bright side facing the Sun has a constant uniform temperature $T_{1}$, while the dark side has a constant uniform temperature $T_{2}$. The orbit radius of the planet is $R_{0}$, the Sun has temperature $T_{s}$, and the radius of the Sun is $R_{s}$. Assume that outer space has zero temperature, and treat all objects as ideal blackbodies.

Question: a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma A T^{4}$.)

Solution:
The intensity of solar radiation at the surface of the sun is $\sigma T_{s}^{4}$, so the intensity at the planet's orbit radius is



$$

I=\sigma T_{s}^{4} \frac{R_{s}^{2}}{R_{0}^{2}}

$$



The area subtended by the planet is $\pi R^{2}$, so



$$

P=\pi \sigma T_{s}^{4} \frac{R^{2} R_{s}^{2}}{R_{0}^{2}}

$$

Final answer: $P=\pi \sigma T_{s}^{4} \frac{R^{2} R_{s}^{2}}{R_{0}^{2}}$
Subfield: Thermodynamics


Question: b. The equilibrium temperature ratio $T_{2} / T_{1}$ depends on the heat transfer rate between the hemispheres. Find the minimum and maximum possible values of $T_{2} / T_{1}$. In each case, what is the wind power $P_{w}$ produced?

Solution:
The minimum value is simply zero; in this case zero heat is transferred to the dark side of the planet. Since no heat is transferred, the heat engine can't run, so $P_{w}=0$. (To show this a bit more carefully, note that the entropy exhausted by the heat engine is $Q_{2} / T_{2} \propto T_{2}^{3}$ by the Stefan-Boltzmann law. In the limit $T_{2} \rightarrow 0$, the entropy out goes to zero, so the entropy in and hence the heat intake also goes to zero.)



The maximum value is $T_{2} / T_{1}=1$. It cannot be any higher by the second law of thermodynamics. In this case, there is no temperature difference, so the heat engine has zero efficiency and $P_{w}=0$. Power $P / 2$ is simply transferred from the bright side to the dark side as heat.

Final answer: The minimum value is simply zero, $P_{w}=0$.
The maximum value is $T_{2} / T_{1}=1$. $P_{w}=0$.
Subfield: Electromagnetism

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1

\end{array}

$$

Extra Supplementary Reading Materials:

Electric Slide



Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium.

Question: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that



$$

v=\mu E

$$



where $E$ is the local electric field and $\mu$ is the charge mobility.

i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives.

Solution:
In the steady state, the current is the same everywhere. Consider the region $(x, x+d x)$. The time it takes for the charge in the second region to leave is $\frac{\mathrm{d} x}{v(x)}$. The amount of charge that leaves is $\rho A \mathrm{~d} x$. The current is thus given by $\rho A v$, so $\rho v$ is constant. Alternatively, one can write this as



$$

v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0

$$



Both forms were accepted.



Final answer: $v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$
Subfield: Electromagnetism


Question: ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.)

Solution:
Let us find an expression for the electric field at position $x$. The position $x$ is effectively in between two uniform sheets of charge density. The sheet on the left has charge density $\int_{0}^{x} \rho \mathrm{d} x+\sigma_{0}$, where $\sigma_{0}$ is the charge density on the left plate, and the sheet on the right has charge density $\int_{x}^{d} \rho \mathrm{d} x+\sigma_{d}$, where $\sigma_{d}$ is the charge density on the left plate. Then, the electric field is given by



$$

E=\sigma_{0} /\left(2 \epsilon_{0}\right)+\int_{0}^{x} \rho /\left(2 \epsilon_{0}\right) \mathrm{d} x-\int_{x}^{d} \rho /\left(2 \epsilon_{0}\right) \mathrm{d} x-\sigma_{d} /\left(2 \epsilon_{0}\right)

$$



Then, by the Fundamental Theorem of Calculus



$$

\frac{d E}{d x}=\frac{\rho}{\epsilon_{0}}

$$







so



$$

\frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}}

$$



This can also be derived from the differential form of Gauss's Law more easily and is known as Poisson's equation.

Final answer: $$

\frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}}

$$
Subfield: Modern Physics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1

\end{array}

$$

Extra Supplementary Reading Materials:

## Strain in the Membrane ${ }^{2}$

<img_4543>


The wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature.



The neuron has "ion pumps" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\epsilon$.

Question: a. Suppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane?

Solution:
One charge layer by itself creates an electric field $E_{1}=Q /(2 \epsilon A)$ in each direction. So the force between the two sides of the membrane $F_{E}=Q E_{1}=Q^{2} /(2 \epsilon A)$.



This electric force is balanced by the spring force $F_{s}=k x$, where $x=d_{0}-d$. Equating these two forces and solving for $d$ gives gives



$$

d=d_{0}-\frac{Q^{2}}{2 \epsilon A k}

$$

Final answer: $d=d_{0}-\frac{Q^{2}}{2 \epsilon A k}$
Subfield: Modern Physics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1

\end{array}

$$

Extra Supplementary Reading Materials:

## Strain in the Membrane ${ }^{2}$

<img_4543>


The wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature.



The neuron has "ion pumps" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\epsilon$.
Context question:
a. Suppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane?
Context answer:
\boxed{$d=d_{0}-\frac{Q^{2}}{2 \epsilon A k}$}


Question: b. Derive an expression for the voltage difference $V$ between the outer and inner surfaces of the membrane in terms of $Q$ and the other parameters given.

Solution:
The electric field inside the membrane (as produced by both the left and right plates) is $E=Q /\left(\epsilon_{0} \kappa A\right)$. So the voltage between them is



$$

V=E d=\frac{Q}{\epsilon A} d

$$



Inserting the expression for $Q$ from part (a) gives



$$

V=\frac{Q}{\epsilon A}\left(d_{0}-\frac{Q^{2}}{2 \epsilon A k}\right)

$$



This equation implies that as the charge $Q$ is increased, the voltage first increases and then decreases again.

Final answer: $V=\frac{Q}{\epsilon A}\left(d_{0}-\frac{Q^{2}}{2 \epsilon A k}\right)$
Subfield: Mechanics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{array}

$$


$$

\begin{array}{lrr}

M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\

L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\

R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\

\lambda_{\max } & = & 500 \mathrm{~nm}

\end{array}

$$

Question: a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible.


i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?

Solution:
During the collision, the block receives impulses from the normal force, friction, and gravity. Since the collision is very short, the impulse due to gravity is negligible. Let $p_{N}$ and $p_{F}$ be the magnitudes of the impulses from the normal force and friction force.



Since the block stays on the ramp after the collision, its final momentum is parallel to the ramp. Then the normal force must completely eliminate the block's initial momentum perpendicular to the ramp, so



$$

p_{N}=m v \cos \theta

$$



Since the block still moves after the collision,



$$

p_{F}=\mu p_{N}

$$



The block's initial momentum parallel to the ramp is $m v \sin \theta$, so



$$

m v \sin \theta-p_{F}=m u

$$



where $u$ is the final speed of the block. Solving for $u$ gives



$$

u=v(\sin \theta-\mu \cos \theta)

$$

Final answer: $u=v(\sin \theta-\mu \cos \theta)$
Subfield: Mechanics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{array}

$$


$$

\begin{array}{lrr}

M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\

L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\

R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\

\lambda_{\max } & = & 500 \mathrm{~nm}

\end{array}

$$

Context question:
a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible.


i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?
Context answer:
\boxed{$u=v(\sin \theta-\mu \cos \theta)$}


Question: ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ?

Solution:
We set $u=0$ to obtain



$$

\mu=\tan \theta

$$



Note that this is simply the no-slip condition for a block resting on an inclined plane! This is because in both cases, equality is achieved when the normal force and maximal friction force sum to a purely vertical force.

Final answer: $\mu=\tan \theta$
Subfield: Mechanics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{array}

$$


$$

\begin{array}{lrr}

M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\

L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\

R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\

\lambda_{\max } & = & 500 \mathrm{~nm}

\end{array}

$$

Context question:
a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible.


i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?
Context answer:
\boxed{$u=v(\sin \theta-\mu \cos \theta)$}


Context question:
ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ?
Context answer:
\boxed{$\mu=\tan \theta$}


Question: b. Now suppose you drop a sphere with mass $m$, radius $R$ and moment of inertia $\beta m R^{2}$ vertically onto the same fixed ramp such that it reaches the ramp with speed $v$.


i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case?

Solution:
If the sphere immediately begins to roll without slipping, we can calculate the frictional impulse independently of the normal impulse. We have



$$

m v \sin \theta-p_{F}=m u

$$



The frictional impulse is responsible for the sphere's rotation, so its angular momentum about its center of mass is $L=p_{F} R$. But we also know that



$$

L=\beta m R^{2} \omega=\beta m R u

$$



Then



$$

p_{F}=\beta m u \text {. }

$$



Substituting into the previous expression gives



$$

m v \sin \theta=(1+\beta) m u \Rightarrow u=\frac{v \sin \theta}{1+\beta}

$$

Final answer: $u=\frac{v \sin \theta}{1+\beta}$
Subfield: Mechanics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{array}

$$


$$

\begin{array}{lrr}

M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\

L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\

R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\

\lambda_{\max } & = & 500 \mathrm{~nm}

\end{array}

$$

Context question:
a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible.


i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?
Context answer:
\boxed{$u=v(\sin \theta-\mu \cos \theta)$}


Context question:
ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ?
Context answer:
\boxed{$\mu=\tan \theta$}


Context question:
b. Now suppose you drop a sphere with mass $m$, radius $R$ and moment of inertia $\beta m R^{2}$ vertically onto the same fixed ramp such that it reaches the ramp with speed $v$.


i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case?
Context answer:
\boxed{$u=\frac{v \sin \theta}{1+\beta}$}


Question: ii. What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision?

Solution:
As in part (a), the normal impulse is $p_{N}=m v \cos \theta$ and the maximal frictional impulse is $p_{F}=\mu p_{N}$. From the previous part, we need



$$

p_{F}=\frac{\beta m v \sin \theta}{1+\beta}

$$



and equating these expressions gives



$$

\mu=\frac{\beta \tan \theta}{1+\beta}

$$

Final answer: $\mu=\frac{\beta \tan \theta}{1+\beta}$