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Условие Отговори Решение

USAPhO 2001, Четвъртфинал — Задача 3

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Всички материали от USAPhO 2001

2001 Preliminary Exam — Open Response Questions · 25 т.

Preliminary Exam

Open Response Questions

4 Questions, 60 minutes

Указания

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 $g=9.8\ N/kg$

Newton’s gravitational constant $G=6.67\times10^{-11}\ N\cdot m^2/kg^2$

Binomial expansion $(1+x)^n\approx1+nx$ for $|x|\ll1$

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. 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 $mv_0r_0=mvr$ , 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

In this problem, please use $g=10\ N/kg$ . A bead of mass $0.050\ kg$ is placed on a long vertical wire which is firmly attached to both the floor and ceiling. As the bead slides on the wire the magnitude of the frictional force between the bead and the wire is a constant $0.20\ N$ . The bead is released from rest at an initial height of $1.47\ m$ above the floor. Assume all collisions with the floor are perfectly elastic and that air resistance can be neglected.

a What is the bead's greatest height after colliding with the floor? [9 т.]

b How much time does it take the bead to slide from its initial height of $1.47\ m$ to the floor? [6 т.]

c How much time does it take the bead to travel from the floor to the height found in a) above? [5 т.]

d What is the total distance the bead travels before coming to rest? [5 т.]

Отговори

Решение

$m=0.050\ kg$ , $mg=(0.050\ kg)(10\ m/s^2)=0.50\ N$ , $h=1.47\ m$ , $f=0.20\ N$ .

a. Use the work energy theorem

W_{nc}=\Delta K+\Delta U\qquad\text{Eq. 3-1}

[2] where $W_{nc}$ is the work done by the non-conservative frictional force, $\Delta K$ is the change in kinetic energy, and $\Delta U$ is the change in potential energy. Let $h'$ = greatest height after colliding with the floor. We will compare energies at the initial height $h$ and the final height $h'$ . The frictional force is always opposite to the direction of motion, so the work done is the negative of the frictional force times the total distance traveled.

W_{nc}=-f(h+h')

[2]

\Delta K=0

[1]

\Delta U=mg(h'-h)

[2] Combining with Eq. 3-1,

-f(h+h')=mg(h'-h)

[1] and solving for $h'$ , $(mg+f)h'=(mg-f)h$ ,

h'=h\left(\frac{mg-f}{mg+f}\right)=(1.47\ m)\left(\frac{0.50\ N-0.20\ N}{0.50\ N+0.20\ N}\right)=0.63\ m.

[1]

b. As the bead moves downward, the frictional force is upward. Choosing the downward (acceleration) direction to be positive and the origin of the coordinate system to be at the point of release, we have $\sum F=ma$ . [1]

mg-f=ma

[1]

a=\frac{mg-f}{m}=\frac{0.50\ N-0.20\ N}{0.050\ kg}=6.0\ m/s^2

[1]

y=y_0+v_0t+\frac12at^2

[1] with $y=h=1.47\ m$ , $y_0=0$ , and $v_0=0$ :

t=\sqrt{\frac{2h}{a}}=\sqrt{\frac{2(1.47\ m)}{6.0\ m/s^2}}=0.70\ s.

[2]

c. In order to find the initial velocity after the bounce, find the final velocity of the bead immediately before the bounce.

v=v_0+at=0+(6.0\ m/s^2)(0.70\ s)=4.2\ m/s\quad\text{(downward)}.

[1] Since the collision with the floor is elastic, the initial velocity of the bead after the bounce is $v_0=-4.2\ m/s$ (upward). [1]

As the bead moves upward $\sum F=ma$ .

The frictional force is downward $mg+f=ma$ [1]

a=\frac{mg+f}{m}=\frac{0.50\ N+0.20\ N}{0.050\ kg}=14.0\ m/s^2.

[1] At the highest point, $v=0$ , and

t=\frac{v-v_0}{a}=\frac{0-(-4.2\ m/s)}{14.0\ m/s^2}=0.30\ s.

[1]

d. Use the work energy theorem $W_{nc}=\Delta K+\Delta U$ . [1] where $W_{nc}$ is the work done by the non-conservative frictional force, $\Delta K$ is the change in kinetic energy, and $\Delta U$ is the change in potential energy. We will compare energies at the initial height $h$ and the final height $0$ . Let $s$ = total distance traveled. The work done by the frictional force is the negative of the frictional force times the total distance traveled.

W_{nc}=-fs

[1]

\Delta K=0,\qquad\Delta U=0-mgh

[1] Combining $-fs=-mgh$ [1] and solving for $s$

s=\frac{mgh}{f}=\frac{(0.50\ N)(1.47\ m)}{0.20\ N}=3.7\ m.

[1]

Теми и трудност

Трудност: неоценена.

Механика · Работа и енергия · Триене · Кинематика

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Оригинал в Архива: Quarterfinal 2001.pdf · официални решения: Quarterfinal 2001 Solutions.pdf