Съдържание

Условие Отговори Решение

USAPhO 2005, Четвъртфинал — Задача 3

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Всички материали от USAPhO 2005

2005 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

Quantity	Value
Gravitational field at the Earth's surface	$g=9.8\ N/kg$
Newton's gravitational constant	$G=6.67\times10^{-11}\ N\cdot m^2/kg^2$
Average Earth-Sun Distance	$1$ Astronomical Unit (A.U.) $=1.5\times10^{11}\ m$
Binomial expansion	$(1+x)^n=1+nx$ for $\vert x\vert \ll1$

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. Where alternate methods are presented, a student should be awarded points from only one method. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.) 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 2.b., if a student wrote down $mvb=mv_pR$ , they should get 10 points credit.

Условие

Compressed spring and block on a rough surface with initial displacement A0

A mass $M$ with weight $Mg=10.00\ N$ is placed at the end of a massless spring with spring constant $100.0\ N/m$ as shown in the above diagram. The coefficients of static and kinetic friction between the mass and the rough horizontal surface are $0.400$ and $0.200$ , respectively. The spring is compressed a distance $A_0=0.180\ m$ from its equilibrium position and released from rest. Find the total distance the mass travels and where it comes to rest with respect to the equilibrium position.

Отговори

Решение

In order for the mass to continue to oscillate back and forth, the spring force at maximum extension must exceed the maximum force of static friction. Letting $A_i$ equal the maximum displacement after the $i$ th half cycle.

The mass stops oscillating when $F|_{x=A_i}<f_{s,max}$ . [2] The spring force has magnitude $F=kx$ . [1] The maximum force of static friction is $f_{s,max}=\mu_sN=\mu_sMg$ . [2] The oscillation stops if $kA_i<\mu_sMg$ . [1] Or numerically,

A_i<\frac{\mu_sMg}{k}=\frac{(0.400)(10.00\ N)}{100.0\ N/m}=0.0400\ m.\qquad (3-1)

[1] Applying the work energy theorem to a single half oscillation,

W_{nc}+K_i+U_i=K_{i+1}+U_{i+1},\qquad (3-2)

[2] where $W_{nc}$ is the work done by non-conservative forces, friction in this case. At the extreme displacements $A_i$ and $A_{i+1}$ , the kinetic energy is zero: at $x=A_i$ and $x=A_{i+1}$ , $K_i=0$ and $K_{i+1}=0$ . [1]

The spring potential energy at $A_i$ and $A_{i+1}$ is $U_i=\frac12kA_i^2$ and $U_{i+1}=\frac12kA_{i+1}^2$ . [1] The work done by kinetic friction is $W_{nc}=-f_kd=-\mu_kMgd$ , [2] where $d$ is the total distance traveled, or $d=A_i+A_{i+1}$ . [1]

W_{nc}=-\mu_kMg(A_i+A_{i+1}).

[1] Substituting into (3-2),

-\mu_kMg(A_i+A_{i+1})+\frac12kA_i^2=\frac12kA_{i+1}^2.

[1]

\mu_kMg(A_i+A_{i+1})=\frac12k(A_i^2-A_{i+1}^2)=\frac12k(A_i+A_{i+1})(A_i-A_{i+1}).

Dividing by $(A_i+A_{i+1})$ , $\mu_kMg=\frac12k(A_i-A_{i+1})$ . Solving for $A_{i+1}$ ,

A_{i+1}=A_i-\frac{2\mu_kMg}{k}.

[2] Numerically,

A_{i+1}=A_i-\frac{2(0.200)(10.00\ N)}{100.0\ N/m}=A_i-0.0400\ m.

Therefore each half cycle the amplitude decreases by $0.040\ m=4.0\ cm$ . [1]

Successive amplitudes (starting with the initial displacement) are $0.180\ m$ , $0.140\ m$ , $0.100\ m$ , $0.060\ m$ , $0.020\ m$ . [2] At $0.020\ m$ the amplitude satisfies condition (3-1) and the mass remains at rest. Since $A_0$ is to the left of the equilibrium position and they alternate, the stopping point is $0.020\ m$ to the left of the equilibrium position. [2]

The total distance traveled is $D=A_0+2A_1+2A_2+2A_3+A_4$ .

D=(0.180\ m)+2(0.140\ m)+2(0.100\ m)+2(0.060\ m)+(0.020\ m)=0.800\ m.

[2]

Теми и трудност

Трудност: неоценена.

Механика · Триене · Механични трептения · Работа и енергия

Предпоставки: Разширена олимпиадна подготовка

Оригинал в Архива: Quarterfinal 2005.pdf · официални решения: Quarterfinal 2005 Solutions.pdf