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Условие Отговори Решение

USAPhO 2002, Четвъртфинал — Задача 2

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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 particle of mass $m$ is constrained to move in one dimension. A force $\mathbf{F}$ acts on the particle. $\mathbf{F}$ always points toward the position labeled $E$ . For example, when the particle is to the left of $E$ , $\mathbf{F}$ points to the right. The magnitude of $\mathbf{F}$ is a constant $F$ except at point $E$ where it is zero.

A horizontal line with a particle m drawn as a black dot to the left of a vertical tick marked E; a grey arrow labelled A points from the tick at E leftwards to above the particle, showing the displacement A.

a. The system is horizontal. $\mathbf{F}$ is the net force acting on the particle. The particle is displaced a distance $A$ from the equilibrium position $E$ and released from rest. What is the period of the motion? [10 т.]

b. The system is vertical. Both the force $\mathbf{F}$ described above and gravity act on the system. The particle is displaced a distance $A$ above the position $E$ and released from rest. $F > mg$ . What is the period of the motion? [15 т.]

Отговори

Решение

2. a. When the particle is released, it begins to move with an acceleration which has magnitude

a = \frac{F}{m}

[2]

and is directed toward position $E$ . Since it starts from rest $\quad A = \tfrac{1}{2}at^2$ [2]

It reaches $E$ in time $t$

t = \sqrt{\frac{2A}{a}} = \sqrt{\frac{2Am}{F}}

[1]

The particle oscillates back and forth between amplitude A on the right and A on the left. [2]

The time for one complete cycle is four times this $\quad T = 4t$ [2]

T = 4\sqrt{\frac{2Am}{F}}

[1]

b. The particle oscillates between amplitude $A$ above point $E$ and $A'$ below. $A \ne A'$ [1]

Let $t_1$ = time it takes the particle to go from amplitude $A$ above to point $E$ and from $E$ up to $A$ .

Let $t_2$ = time it takes the particle to go from point $E$ to amplitude $A'$ below and from $A'$ up to $E$ .

The period is $\quad T = 2t_1 + 2t_2$ [2]

When the particle is above point E, the net force acting on it is

ma = \sum F = F + mg \quad \text{downward.}

[1]

The magnitude of the acceleration is $\quad a = \dfrac{F + mg}{m}$ [1]

Starting from rest the particle travels a distance $A$ given by $\quad A = \tfrac{1}{2}at_1^{\,2}$ [1]

Solving for $t_1$

t_1 = \sqrt{\frac{2A}{a}} = \sqrt{\frac{2Am}{F + mg}}

[1]

The particle passes through E with velocity

v_E = at_1 = \left(\frac{F + mg}{m}\right)\sqrt{\frac{2Am}{F + mg}}

[1]

When the particle is below point E, the net force acting on it is

ma = \sum F = F - mg \quad \text{upward}

[1]

The magnitude of the acceleration is $\quad a = \dfrac{F - mg}{m}$ [1]

It reaches maximum downward displacement $A'$ when $v = 0$ . [1]

v = 0 = v_E - at_2

[1]

Solving for $t_2$

t_2 = \frac{v_E}{a} = \left(\frac{F + mg}{m}\right)\sqrt{\frac{2Am}{F + mg}}\left(\frac{m}{F - mg}\right) = \left(\frac{F + mg}{F - mg}\right)\sqrt{\frac{2Am}{F + mg}}

[1]

T = 2t_1 + 2t_2 = 2\sqrt{\frac{2Am}{F + mg}} + 2\left(\frac{F + mg}{F - mg}\right)\sqrt{\frac{2Am}{F + mg}}

[1]

T = 2\sqrt{\frac{2Am}{F + mg}}\left(1 + \frac{F + mg}{F - mg}\right) = \left(\frac{4F}{F - mg}\right)\sqrt{\frac{2Am}{F + mg}}

[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