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Условие Отговори Решение

USAPhO 2002, Четвъртфинал — Задача 1

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Всички материали от USAPhO 2002

AAPT/AIP United States Physics Team 2002 — Preliminary Exam, Open Response Questions · 25 т.

Указания

4 Questions, 60 minutes

INSTRUCTIONS

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 $g = 9.8\ \mathrm{N/kg}$

Newton’s gravitational constant $G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}$

Binomial expansion $(1 + x)^n \approx 1 + nx$ for $|x| \ll 1$

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

Disk	$\frac{1}{2} MR^2$
Sphere	$\frac{2}{5} MR^2$
Rod	$\frac{1}{12} ML^2$

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 $mgH = Fh$ , they should get 9 points credit. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

A rocket of mass $m$ is launched straight up, reaches a maximum height $H$ above the ground, and begins its vertical descent. The rocket is in free fall until it falls to a height $h$ . At that point, the rocket’s engines are fired and provide a continuous upward thrust $F$ . You may neglect the following: air resistance; any horizontal motion of the rocket; and any change in the mass of the rocket. Express your answers in terms of only the variables given ( $H$ , $F$ , $m$ ) and, if necessary, the gravitational field at the Earth’s surface $g$ and the radius of the Earth $R_E$ .

a. $H$ is small enough that the force of gravity is constant during the flight of the rocket. At what height $h$ above the surface of the Earth should the engines be fired so that the rocket lands safely with zero velocity? [10 т.]

b. For what values of $F$ is a zero velocity landing impossible? [5 т.]

c. $H$ is large enough that the force of gravity is not constant during the flight of the rocket. At what height $h$ above the surface of the Earth should the engines be fired so that the rocket lands safely with zero velocity? [10 т.]

Отговори

Решение

1. a. Using the work-energy theorem $\mathrm{PE}_i + \mathrm{KE}_i + W = \mathrm{PE}_f + \mathrm{KE}_f$ [1]

“i” refers to the point of maximum height $H$ and “f” refers to the Earth’s surface. At both points the kinetic energy is zero

$\mathrm{KE}_i = 0 \qquad \mathrm{KE}_f = 0.$ [2]

Setting the potential energy at the Earth’s surface equal to zero

$\mathrm{PE}_i = mgH \qquad \mathrm{PE}_f = 0.$ [2]

The work done by the constant force F $\quad W = \vec{F}\cdot\vec{d}$ [1]

$\vec{F}$ is upward, displacement $\vec{d}$ is downward and has magnitude $h$ . $\quad W = -Fh.$ [2]

Inserting these into the work energy equation $\quad mgH - Fh = 0$ [1]

h = \frac{mg}{F}H.

[1]

b. A zero velocity solution is impossible if $h \ge H$ . Then [2]

H \le \frac{mg}{F}H,

[1]

F \le mg.

[2]

$\{$ Minus one point if $<$ instead of $\le$ . $\}$ $\{$ A check of the equality shows that for $h = H$ , $F = mg$ . So at the highest point its velocity is zero and the net force is zero. It just stays there and will not reach ground with zero velocity. This argument is not required for full credit. $\}$

c. Using the work-energy theorem $\mathrm{PE}_i + \mathrm{KE}_i + W = \mathrm{PE}_f + \mathrm{KE}_f$

“i” refers to the point of maximum height $H$ and “f” refers to the Earth’s surface. At both points the kinetic energy is zero

$\mathrm{KE}_i = 0 \qquad \mathrm{KE}_f = 0.$ [1]

Here the gravitational potential energy is $\quad PE_i = -\dfrac{GM_E m}{R_E + H} \qquad PE_f = -\dfrac{GM_E m}{R_E}.$ [2]

Where $M_E$ is the mass of the Earth. The work done is $\quad W = -Fh$ [2]

(see above). Inserting these into the work energy equation

-\frac{GM_E m}{R_E + H} - Fh = -\frac{GM_E m}{R_E}

[1]

solving for $h$

h = \frac{GM_E m}{F}\left(\frac{1}{R_E} - \frac{1}{R_E + H}\right) = \frac{GM_E m}{F}\left(\frac{R_E + H - R_E}{R_E(R_E + H)}\right) = \frac{GM_E mH}{FR_E(R_E + H)}.

[1]

Answer must be expressed in terms of $m$ , $H$ , $F$ , $g$ , and $R_E$ . Need to eliminate $G$ and $M_E$ .

g = \frac{GM_E}{R_E^{\,2}}

[1]

Solving for $\quad GM_E = gR_E^{\,2}.$ [1]

Inserting this into the equation for $h$

h = \frac{\left(gR_E^{\,2}\right)mH}{FR_E(R_E + H)} = \frac{mgHR_E}{F(R_E + H)} = \frac{mg}{F}H\left(\frac{1}{1 + H/R_E}\right).

[1]

Автори

The 2002 Preliminary Examination open response questions were written by the coaches of the United States Physics Team. The coaches are: Academic Director, Mary Mogge – Professor of Physics at California State Polytechnic University, Pomona, CA; Senior Coach, Leaf Turner – Physicist in the Theoretical Division of Los Alamos National Laboratory, Los Alamos, NM; Warren Turner – Physics Teacher at Massachusetts Academy of Math and Science, Worcester, MA; Robert Shurtz – Physics Teacher at the Hawken School, Gates Mills, OH. Questions 1 and 3 were suggested by Warren Turner, Question 2 by Mary Mogge, and Question 4 by Leaf Turner.

Оригинал в Архива: Quarterfinal 2002.pdf · официални решения: Quarterfinal 2002 Solutions.pdf