Съдържание

Условие Отговори Решение

USAPhO 2005, Четвъртфинал — Задача 1

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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

A shell is shot from the edge of a $4.00\ m$ high cliff with initial velocity $v_i$ at an angle of $45.0^\circ$ above the horizontal. It lands $18.0\ m$ from the base of the cliff as shown in the accompanying diagram.

Shell launched at 45 degrees from a 4.00 m cliff and landing 18.0 m from its base

a What is the initial speed $v_i$ of the shell? [13 т.]

b A second shell is launched with the same initial velocity. At the top of its trajectory it explodes into two fragments. One fragment of mass $m_1$ has zero speed immediately after the explosion and falls vertically. The other fragment of mass $m_2$ lands $30.0\ m$ from the base of the cliff. Find the ratio $m_1/m_2$ . [12 т.]

Отговори

Решение

a. Setting the coordinate origin at the top of the cliff, the horizontal position is given by

x=v_{ix}t=(v_i\cos\theta)t=(v_i\cos45.0^\circ)t=\frac{v_it}{\sqrt2}.\qquad (1-1)

[3] Taking up to be positive, the vertical position is given by

y=v_{iy}t-\frac12gt^2=(v_i\sin45.0^\circ)t-\frac12gt^2=\frac{v_it}{\sqrt2}-\frac12gt^2.\qquad (1-2)

[4] Solving (1-1) for $t$ ,

t=\frac{x\sqrt2}{v_i}.\qquad (1-3)

[1] Substituting into (1-2),

y=\frac{v_i}{\sqrt2}\left(\frac{x\sqrt2}{v_i}\right)-\frac12g\left(\frac{x\sqrt2}{v_i}\right)^2=x-g\frac{x^2}{v_i^2}.

[2] Solving for $v_i$ , $v_i=\sqrt{gx^2/(x-y)}$ . [1] The shell lands at $x=18.0\ m$ and $y=-4.00\ m$ , and

v_i=\sqrt{\frac{(9.8\ m/s^2)(18.0\ m)^2}{18.0\ m-(-4.0\ m)}}=12.0\ m/s.

[2]

b. The $y$ -component of the velocity is $v_y=v_{iy}-gt=v_i/\sqrt2-gt$ . [2] At its highest point, $v_y=0$ and

t_h=\frac{v_i}{g\sqrt2}.\qquad (1-4)

[2] The shell's $x$ position is

x=\frac{v_it_h}{\sqrt2}=\frac{v_i}{\sqrt2}\left(\frac{v_i}{g\sqrt2}\right)=\frac{v_i^2}{2g}=\frac{(12.0\ m/s)^2}{2(9.8\ m/s^2)}=7.36\ m.

[1]

METHOD I: The explosion is an internal force and the motion of the shell's center of mass is unchanged. [2]

(Note: Since fragment 1 has zero velocity after the explosion, the impulse is totally in the $x$ -direction and the fragments land at the same time.)

Let $x_i$ be the landing position of fragment $m_i$ . Fragment 1 falls straight down, so $x_1=7.36\ m$ while $x_2=30.0\ m$ . [1] The defining equation of the center of mass of the two fragments is

(m_1+m_2)x_{cm}=m_1x_1+m_2x_2.

[2]

\frac{m_1}{m_2}=\frac{x_2-x_{cm}}{x_{cm}-x_1}=\frac{30.0\ m-18.0\ m}{18.0\ m-7.4\ m}=1.13.

[2]

METHOD II: The explosion is an internal force and the total momentum of the shell is unchanged. [1]

(Note: Since fragment 1 has zero velocity after the explosion, the impulse is totally in the $x$ -direction and the $y$ -component of momentum is unaffected.)

For the $x$ -component, momentum conservation gives

(m_1+m_2)v_{ix}=m_1(0)+m_2v_2=m_2v_2.

[1] The velocity of $m_2$ after the explosion is

v_2=\frac{(m_1+m_2)v_{ix}}{m_2}=\left(1+\frac{m_1}{m_2}\right)\frac{v_i}{\sqrt2}.

[1] The fragment's position after the explosion is

x_2=x_h+v_2t_f=x_h+\left(1+\frac{m_1}{m_2}\right)\frac{v_it_f}{\sqrt2}.\qquad (1-5)

[1] Here $t_f$ is the time it take the fragment to fall from its highest point to the ground. Since the $y$ -component of velocity is unchanged by the explosion, the time to fall is the total time (1-3) minus the time to reach the highest point (1-4):

t_f=t-t_h=\frac{x\sqrt2}{v_i}-\frac{v_i}{g\sqrt2}=\frac{(18.0\ m)\sqrt2}{12.0\ m/s}-\frac{12.0\ m/s}{(9.8\ m/s^2)\sqrt2}=1.255\ s.

[2]

(The time can also be obtained by first finding the $y$ -displacement at the highest point and then finding the time to fall from that point to the ground. Using either method a time of $1.26\ s$ should get 2 points credit.)

Solving (1-5) for the mass ratio,

\frac{m_1}{m_2}=\frac{(x_2-x_h)\sqrt2}{v_it_f}-1=\frac{(30.0\ m-7.36\ m)\sqrt2}{(12.0\ m/s)(1.255\ s)}-1=1.13.

[1]

Теми и трудност

Трудност: неоценена.

Механика · Движение на хвърлено тяло · Импулс и сблъсъци · Център на масите

Предпоставки: Разширена олимпиадна подготовка

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