Съдържание

Условие Отговори Решение

USAPhO 2004, Четвъртфинал — Задача 4

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Всички материали от USAPhO 2004

AAPT AIP United States Physics Team 2004 — 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

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

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

Disk	$^1/_2\,MR^2$
Sphere	$^2/_5\,MR^2$
Rod	$^1/_{12}\,ML^2$

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 4.a., if a student wrote down $\omega = (Mv\alpha L)/(2mL^2)$ , 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

A dumbbell of length $2L$ is initially at rest on a horizontal surface. The accompanying diagram shows a view looking down at the surface. Both cubes at the ends of the dumbbell have mass $m$ . The connecting rod is massless. An object with mass $M$ and velocity $v$ perpendicular to the rod slides across the surface and strikes the dumbbell a distance $\alpha L$ from its center. The collision is perfectly elastic. There is no friction between any of the objects and the surface. Assume the masses $M$ and $m$ can be treated as point masses. After the collision, mass $M$ is at rest.

Top view: a vertical rod with a cube m at each end; dashed dimension arrows L and L from the center to each end. A cube M on the left moves right with velocity v toward the rod, striking it a distance αL below the center.

a. Find $M$ in terms of some or all of the quantities $\alpha$ , $L$ , $v$ , and $m$ . [15 т.]

b. What is $M$ if $\alpha = 1$ , i.e., $M$ strikes $m$ at the end of the rod? [5 т.]

c. What is $M$ if $\alpha = 0$ , i.e., $M$ strikes at the center of the rod? [5 т.]

Note: Parts (b) and (c) can be answered without first doing Part (a).

Отговори

Решение

4. Before the collision the dumbbell is at rest and mass $M$ moves with velocity $v$ . After the collision $M$ is at rest and the dumbbell is both translating and rotating. Let $v_{cm}$ equal the translational velocity of the dumbbell center of mass and let $\omega$ equal the angular velocity of the dumbbell about its center of mass.

The moment of inertia of the dumbbell about its center of mass is $I = 2mL^2$ [2]

Angular momentum is conserved. The initial angular momentum of $M$ about the dumbbell’s center of mass equals the final angular momentum of the dumbbell about its center of mass.

$Mv(\alpha L) = I\omega$ [3]

Solving for $\omega$ $\omega = \frac{Mv\alpha L}{I} = \frac{Mv\alpha L}{2mL^2} = \frac{Mv\alpha}{2mL}$ [1]

Linear momentum is conserved. $Mv = 2mv_{cm}$ [3]

Solving for the center of mass velocity $v_{cm} = \frac{Mv}{2m}$ . [1]

The collision is elastic, so kinetic energy is conserved.

$\frac{1}{2}Mv^2 = \frac{1}{2}(2m)v_{cm}^{\,2} + \frac{1}{2}I\omega^2$ [3]

Substituting into this equation $Mv^2 = (2m)\left(\frac{Mv}{2m}\right)^2 + (2mL^2)\left(\frac{Mv\alpha}{2mL}\right)^2$ [1]

Simplifying $1 = \frac{M}{2m} + \frac{M\alpha^2}{2m} = \frac{M}{2m}(1 + \alpha^2)$

Therefore $M = \frac{2m}{1 + \alpha^2}$ [1]

b. If $\alpha = 1$ $M = \frac{2m}{1 + 1} = m$ [5]

(Immediately after the collision the lower mass $m$ has speed $v$ . $M$ and the other $m$ are at rest. This conserves mechanical energy, linear momentum, and angular momentum.)

c. If $\alpha = 0$ $M = \frac{2m}{1 + 0} = 2m$ [5]

(Immediately after the collision both masses $m$ have speed $v$ . This conserves mechanical energy and linear momentum. Angular momentum about the center of mass is zero in this case.)

Оригинал в Архива: Quarterfinal 2004.pdf · официални решения: Quarterfinal 2004 Solutions.pdf