Съдържание

Условие Отговори Решение

USAPhO 2005, Четвъртфинал — Задача 2. “Hit the planet, win a prize.”

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2005

2005 Preliminary Exam — Open Response Questions · 25 т.

Указания

4 Questions, 60 minutes

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. Cell phones may not be used during the exam or while the exam papers are present. 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.
Good luck!

Possibly Useful Information

Quantity	Value
Gravitational field at the Earth's surface	$g=9.8\ N/kg$
Newton's gravitational constant	$G=6.67\times10^{-11}\ N\cdot m^2/kg^2$
Average Earth-Sun Distance	$1$ Astronomical Unit (A.U.) $=1.5\times10^{11}\ m$
Binomial expansion	$(1+x)^n=1+nx$ for $\vert x\vert \ll1$

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. Where alternate methods are presented, a student should be awarded points from only one method. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.) 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 2.b., if a student wrote down $mvb=mv_pR$ , they should get 10 points credit.

Условие

A space probe is sent toward a distant planet which has mass $M$ and radius $R$ . When the probe is very, very far from the planet, the probe has speed $v$ and is headed in a direction such that, if undeflected, it would pass a distance $b>R$ from the planet's center. See the accompanying diagram.

Incoming probe with speed v and impact parameter b relative to a planet of mass M and radius R

a Assuming that the probe hits the planet, what is the probe's speed just before it hits? [10 т.]

b What is the maximum initial speed the probe can have and still hit the planet? [15 т.]

Отговори

Решение

a. Using energy conservation, the total mechanical energy when the probe is an infinite distance from the planet equals the total mechanical energy just before the probe hits:

K_\infty+U_\infty=K_p+U_p.

[2] Setting $U_\infty=0$ , [1] we have $U_p=-GMm/R$ , [2] where $m$ is the mass of the probe. The probe's kinetic energy is $K_\infty=\frac12mv^2$ and $K_p=\frac12mv_p^2$ , [2] where $v_p$ is the probe's speed just before it hits the surface. Combining,

\frac12mv^2=\frac12mv_p^2-\frac{GMm}{R}.\qquad (2-1)

[1] Solving for $v_p$ , we have $v_p=\sqrt{v^2+2GM/R}$ . [2]

b. A central force, such as the universal gravitational force, causes no torque. Therefore angular momentum is conserved: $L_\infty=L_p$ . [2] Angular momentum is given by

\vec L=\vec r\times\vec p=\vec r\times m\vec v.\qquad (2-2)

[2] Conservation gives $\vec r_\infty\times m\vec v=\vec r_p\times m\vec v_p$ . [1] When the probe is very far away,

|\vec r_\infty\times m\vec v|=mvr_\infty\sin\theta=mvb.\qquad (2-3)

[2] When the probe has the largest speed it can have and still hit the planet, it will come in tangential to the surface, or perpendicular to the radius.

|\vec r_p\times m\vec v_p|=mv_pR.\qquad (2-4)

[2] Combining (2-2), (2-3), and (2-4), $mvb=mv_pR$ . [1] Solving for $v_p$ ,

v_p=v\frac bR.\qquad (2-5)

[1] Substituting (2-5) into (2-1),

\frac12mv^2=\frac12m\left(v\frac bR\right)^2-\frac{GMm}{R}.

[2]

v^2\left[\left(\frac bR\right)^2-1\right]=\frac{2GM}{R},

v=\sqrt{\frac{2GM}{R[(b/R)^2-1]}}=\sqrt{\frac{2GMR}{b^2-R^2}}.

[2]

Теми и трудност

Трудност: неоценена.

Механика · Гравитация · Момент на импулса · Работа и енергия

Предпоставки: Разширена олимпиадна подготовка

Оригинал в Архива: Quarterfinal 2005.pdf · официални решения: Quarterfinal 2005 Solutions.pdf