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OPhO 2020, Открит кръг — Задача 8. Angel Coaster

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Всички материали от OPhO 2020

2020 Online Physics Olympiad (OPhO): Open Contest v1.5 · 25 май 2020 г.

2020 Online Physics Olympiad (OPhO): Open Contest v1.5

OPhO Lead Organizers and Writers: Adithya Balachandran, Eddie Chen, Ashmit Dutta, Ethan Hu, Viraj Jayam, Rishab Parthasarathy, Kushal Thaman, Max Wang, QiLin Xue

OPhO Website Developer: Ben Oommen

Advisors: Dr. Branislav Kisačanin, Kye Shi, Kevin Zhou

Указания

If you wish to request a clarification, please use this form. To see all clarifications, see this document.

Use $g = 9.81\ \mathrm{m/s^2}$ in this contest. See the constants sheet on the following page for other constants.
This test contains 55 short answer questions. Each problem will have three possible attempts.
The weight of each question depends on our scoring system found here. Put simply, the later questions are worth more, and the overall amount of points from a certain question decreases with the number of attempts and days that you take to solve a problem as well as the number of teams who solve it. This means that your score decreases with the number of tries and days you take to solve a given problem.
Any team member is able to submit an attempt. Choosing to split up the work or doing each problem together is up to you. Note that after you have submitted an attempt, your teammates must refresh their page before they are able to see it.
Answers should contain at least three significant figures, unless otherwise specified. All answers within the 1% range will be accepted.
When submitting a response using scientific notation, please use exponential form. In other words, if your answer to a problem is $A \times 10^B$ , please type $AeB$ into the submission portal.
A standard scientific or graphing handheld calculator may be used. Technology and computer algebra systems like Wolfram Alpha or the one in the TI nSpire will not be needed. Attempts to use these tools will be classified as cheating.
You are allowed to use Wikipedia or books in this exam. Asking for help on online forums or your teachers will be considered cheating and may result in a possible ban from future competitions.
Top scorers from this contest will qualify to compete in the Online Physics Olympiad Invitational Contest, which is an olympiad-style exam. More information will be provided to invitational qualifiers after the end of the Open Contest.
In general, answer in SI units (meter, second, kilogram, watt) unless otherwise specified. Please answer all questions in degrees unless otherwise specified.
If the question asks to give your answer as a percent and your answer comes out to be “ $x\%$ ”, please input the value $x$ into the submission form.
Do not put letters in your answer on the submission portal! If your answer is “ $x$ meters” , input only the value $x$ into the submission portal.
Do not communicate information to anyone else apart from your team-members before the exam ends on May 29, 2020 at 11:59 PM UTC.

List of Constants

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, $R = 8.31\ \mathrm{J/(mol \cdot K)}$
Boltzmann’s constant, $k_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e = 1.60 \cdot 10^{-19}$
1 electron volt, $1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}$
Speed of light, $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$
Universal Gravitational constant,
$G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}$
Acceleration due to gravity, $g = 9.81\ \mathrm{m/s^2}$
1 unified atomic mass unit,
$1\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}$
Planck’s constant,
$h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}$
Permittivity of free space,
$\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$
Coulomb’s law constant,
$k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ \mathrm{(N \cdot m^2)/C^2}$
Permeability of free space, $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}$
Magnetic constant,
$\frac{\mu_0}{4\pi} = 1 \cdot 10^{-7}\ \mathrm{(T \cdot m)/A}$
1 atmospheric pressure,
$1\ \mathrm{atm} = 1.01 \cdot 10^{5}\ \mathrm{N/m^2} = 1.01 \cdot 10^{5}\ \mathrm{Pa}$
Wien’s displacement constant, $b = 2.9 \cdot 10^{-3}\ \mathrm{m \cdot K}$
Stefan-Boltzmann constant,
$\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}$

Условие

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 $M = 5\ \mathrm{kg}$ .

A frictionless track: ball m at the top of a curved descent of height h, ball M at rest at the bottom of the curve, followed by a vertical circular loop of radius R on the flat track.

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 $3R$ , 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).

Отговори

Решение

Solution: The velocity of $m$ when it gets to the bottom of the track will be given by

v_b = \sqrt{2gh}.

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

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

mv_b = mv_m + Mv_M

By coefficient of restitution,

v_M - v_m = ev_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 \quad \square

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 + 2gR.

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 \implies v_t = \sqrt{gR}.

Substituting these results into our conservation equation gives us

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

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

mv_b = (M+m)v_f \implies v_f = \frac{mv_b}{M+m}

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

\begin{aligned} \frac{m^2}{(M+m)^2}\left(\sqrt{2gh}\right)^2 &= 5gR \\ \frac{2m^2gh}{(M+m)^2} &= 5gR \\ \frac{6m^2gR}{(M+m)^2} &= 5gR \\ \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} = \boxed{2.37\ \mathrm{m^2}}

Оригинал в Архива: open.pdf · официални решения: open_sol.pdf