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Условие Отговори Решение

USAPhO 2001, Четвъртфинал — Задача 2

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Всички материали от USAPhO 2001

2001 Preliminary Exam — Open Response Questions · 25 т.

Preliminary Exam

Open Response Questions

4 Questions, 60 minutes

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example, Doe, Jamie / Q1 – 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. You may not use any tables, books, or collections of formulas.
Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty.

Possibly Useful Information

Gravitational field at the Earth’s surface $g=9.8\ N/kg$

Newton’s gravitational constant $G=6.67\times10^{-11}\ N\cdot m^2/kg^2$

Binomial expansion $(1+x)^n\approx1+nx$ for $|x|\ll1$

Moment of Inertia about Center of Mass – Uniform Object (will not be provided on the second screening exam)

Object	Moment of Inertia
Disk	$\frac12MR^2$
Sphere	$\frac25MR^2$
Rod	$\frac1{12}ML^2$

Solutions to Problems

Any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets at the right margin. You may further break down the listed points into one point increments. If it is clear they have done an intermediate step, they should get credit for it even if they have not presented it. For example, in 1.a., if a student wrote down $mv_0r_0=mvr$ , they should get 6 points credit. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

Two twin asteroids of identical mass $M$ are located a distance $r$ apart. A probe of mass $m$ (with $m\ll M$ ) is released from rest a distance $A$ (with $A\ll r$ ) from the center of mass of the two-body asteroid system. The probe is an equal distance from both asteroids. Assume that the probe does not affect the separation distance $r$ and that all other objects in the universe are too far away to exert a significant gravitational force. The asteroids must be orbiting their common center of mass in order to maintain a constant distance, but ignore any effects of orbiting in this problem. For both parts, clearly justify any approximations that you are making and give your answers in terms of $r$ , $A$ , $G$ , $M$ , and $m$ .

a How long will it take for the probe to reach the center of mass? [17 т.]

b What will be the speed of the probe as it passes through the center of mass of the asteroid system? [8 т.]

Отговори

Решение

(If a student's only error is using $r$ instead of $r/2$ as the asteroid to center of mass distance, deduct only one point for part a) and one point for part b).)

a. Let $x$ = the distance of the probe from the center of mass of the asteroid system. The diagram below is not to scale. The distance $x$ has been greatly exaggerated. The probe is released from $A=x_{max}$ . Diagram [3]

[Probe displaced from the midpoint between two asteroids and the two gravitational forces]

The magnitude of the force $F$ that each asteroid exerts on the probe is

F=\frac{GMm}{(r/2)^2+x^2}.\qquad\text{Eq. 2-1}

[2]

The total force on the probe is in the negative $x$ -direction and equal to [1]

F_x=-2F\sin\theta\qquad\text{Eq. 2-2}

[1]

with

\sin\theta=\frac{x}{\sqrt{(r/2)^2+x^2}}\qquad\text{Eq. 2-3}

[1]

combining Eqs. 2-1, 2-2, 2-3

F_x=-2G\frac{Mm}{(r/2)^2+x^2}\frac{x}{\sqrt{(r/2)^2+x^2}}=-\frac{2GMmx}{\left((r/2)^2+x^2\right)^{3/2}}

[1]

Since $x\le A\ll r$ ,

\left((r/2)^2+x^2\right)^{3/2}=\left((r/2)^2\right)^{3/2}\left(1+(2x/r)^2\right)^{3/2}\approx(r/2)^3.

[1]

With this approximation, $F_x$ becomes

F_x\approx-\frac{16GMm}{r^3}x

[1]

Comparing this with the defining equation of simple harmonic motion $F=-kx$ [1]

(Students should be given equivalent credit for a valid method using calculus.)

The motion is approximately simple harmonic with

k=\frac{16GMm}{r^3}\qquad\text{Eq. 2-4}

[1]

The period of simple harmonic motion is

T=2\pi\sqrt{\frac{m}{k}}\qquad\text{Eq. 2-5}

[1]

Combining Eqs. 2-4 and 2-5

T=2\pi\sqrt{\frac{m}{(16GMm/r^3)}}=\frac{2\pi}{4}\sqrt{\frac{r^3}{GM}}

[1]

The time $t$ for the probe to reach the center of mass of the asteroid system is

t=\frac{T}{4}

[1]

t=\frac{2\pi}{16}\sqrt{\frac{r^3}{GM}}=\frac{\pi}{8}\sqrt{\frac{r^3}{GM}}

[1]

b. The center of mass of the asteroid system is the equilibrium position for the simple harmonic motion. The probe's velocity as it passes through the equilibrium position is

v_{max}=\omega x_{max}=\omega A.

[2]

\omega=\frac{2\pi}{T}

[2]

\omega=\frac{2\pi}{(2\pi/4)\sqrt{r^3/GM}}=4\sqrt{\frac{GM}{r^3}}.

[2] Then

v_{max}=4A\sqrt{\frac{GM}{r^3}}.

[2]

Probe displaced from the midpoint between two asteroids and the two gravitational forces

Теми и трудност

Трудност: неоценена.

Механика · Гравитация · Механични трептения

Предпоставки: Разширена олимпиадна подготовка

Оригинал в Архива: Quarterfinal 2001.pdf · официални решения: Quarterfinal 2001 Solutions.pdf