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Условие Отговори Решение

USAPhO 2002, Четвъртфинал — Задача 3

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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

An air track glider is fitted with a launcher that fires a projectile with velocity $v_{\mathrm{o}}$ perpendicular to the glider. The launch point LP is marked on the glider, as shown in the diagrams. The glider is placed on a very long air track, which is inclined at an angle $\theta$ to the horizontal. With the air turned on, the system can be considered frictionless. With the air turned off, the coefficient of friction between the track and glider is $\mu$ . The projectile is launched at the instant the glider has velocity $v$ down the track and lands a distance $x$ down the track. While the projectile is in the air, the point LP moves a distance $x_{\mathrm{LP}}$ . What is $\Delta x = x - x_{\mathrm{LP}}$ , if:

The glider (a rectangle with the launcher tube and launch point LP) sitting on a straight track inclined downward to the right; the angle θ between the track and a horizontal line is marked at the lower right. The lower edge of the close-up drawing of the glider (label LP) shows at the top right.

Close-up of the glider: a rectangular block labelled LP with a launcher tube on top containing a small circle; a grey arrow labelled v_o points straight up from the tube.

a. the air is off? [20 т.]

b. the air is on? [5 т.]

Отговори

Решение

3. a. Selecting the $x$ -axis parallel to the inclined plane and $y$ -axis perpendicular to it, the forces acting on the glider are shown as components to the right. Since the glider has no acceleration in the $y$ -direction (perpendicular to the plane),

F_N = mg\cos\theta

[2]

The acceleration in the $x$ -direction can be obtained from Newton’s second law. $\quad \sum F = ma$ [2]

ma_x = mg\sin\theta - f_k

[1]

Subbing in for friction $\quad f_k = \mu F_N = \mu mg\cos\theta$ [1]

a_x = g\sin\theta - \mu g\cos\theta.

[1]

The glider has initial velocity $v$ at time $t = 0$ . The displacement of the glider and point LP at any time is given by

x_{LP} = vt + \tfrac{1}{2}a_x t^2

[1]

x_{LP} = vt + \tfrac{1}{2}(g\sin\theta - \mu g\cos\theta)t^2

[1]

The only force acting on the projectile is gravity. The projectile moves in two dimensions both parallel to the plane and perpendicular to it. The projectile has acceleration components

$a_x = g\sin\theta \qquad a_y = -g\cos\theta$ [2]

and initial velocity components $\quad v_x = v \qquad v_y = v_o$ [2]

Its $x$ and $y$ displacements are given by $\quad x = vt + \tfrac{1}{2}g\sin\theta t^2 \qquad y = v_o t - \tfrac{1}{2}g\cos\theta t^2$ [2]

The $x$ -displacement of the projectile relative to point LP is

\Delta x = x - x_{LP} = vt + \tfrac{1}{2}g\sin\theta t^2 - \left[vt + \tfrac{1}{2}(g\sin\theta - \mu g\cos\theta)t^2\right] = \tfrac{1}{2}\mu g\cos\theta t^2

[2]

When the projectile lands, $\quad y = 0$ [1]

0 = v_o t - \tfrac{1}{2}g\cos\theta t^2

Solving for $t$ and assuming $t \ne 0$

t = \frac{2v_o}{g\cos\theta}

[1]

\Delta x = \tfrac{1}{2}\mu g\cos\theta\left(\frac{2v_o}{g\cos\theta}\right)^2 = \frac{2\mu v_o^{\,2}}{g\cos\theta}

[1]

b. If the air is on the system can be assumed to be frictionless, $\mu = 0$ . [1]

\Delta x = 0

[4]

The projectile lands at LP.

Force components on the glider: two crossed arrows; one along the plane with ends labelled f_k=μF_N (up the slope) and mg sin θ (down the slope), the other perpendicular to the plane with ends labelled F_N (outward) and mg cos θ (into the plane); the angle θ is marked below.

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