Съдържание

Условие Решение

USAPhO 2006, Четвъртфинал — Задача 2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2006

Preliminary Exam — Open Response Questions · 25 т.

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

#1 – P. 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!
In order to maintain exam security, do not communicate any information about the problems or the solutions on this exam (or on the multiple-choice exam) until after February 13th.

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·m $^2$ /kg $^2$
Average Earth-Sun Distance	1 Astronomical Unit (A.U.) $=1.5\times10^{11}$ m
Binomial expansion	$(1+x)^n\simeq1+nx$ for $\lvert x\rvert\ll1$

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

Each question is worth 25 points. Any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets in the right margin after the equation number. 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. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

A small hard block of mass $3m$ is suspended from a thread of length $L$ . A second block of mass $m$ is located on an incline, originally at rest, a height $y$ above the level of the large mass. When the smaller block is released it slides, without friction, down a ramp, and then collides elastically with the larger block. The large block swings around so that the tension in the string just barely drops to zero at the top of the loop. The small block slides back up the ramp, rising to a maximum vertical height $h$ .

Ramp with mass m at height y and hanging mass3m on lengthL; dashed loop trajectory.

a) Find $h$ in terms of $y$ .

b) Find $L$ in terms of $y$ .

Решение

(a) (13 points)

The speed of the small block just before it collides with the large block, $v_0$ , is found from conservation of mechanical energy:

K_f+U_f=K_i+U_i\qquad\text{(2-1)}

\frac12mv_0^2+0=0+mgy\qquad\text{(2-2)}

v_0^2=2gy\qquad\text{(2-3)}

[3 pt]

The speed of the large block just after the collision, $v_2$ , is found from conservation of momentum.

3mv_2=mv_0+mv_1\qquad\text{(2-4)}

3v_2=v_0+v_1\qquad\text{(2-5)}

[2 pt]

where $v_1$ is the speed of the small block just after the collision.

This collision is elastic, so kinetic energy is conserved, or

\frac123mv_2^2=\frac12mv_0^2-\frac12mv_1^2\qquad\text{(2-6)}

3v_2^2=v_0^2-v_1^2\qquad\text{(2-7)}

[2 pt]

Dividing Eq. 2-7 by Eq. 2-5 yields

v_2=v_0-v_1\qquad\text{(2-8)}

which can be combined with Eq. 2-5 to give

v_1=v_0/2\qquad\text{(2-9)}

v_2=v_0/2\qquad\text{(2-10)}

[3 pt]

The speed of the small block as it starts back up the path, $v_1$ , is related to the height that it rises by conservation of mechanical energy (Eq. 2-1), this can be combined with Eq. 2-9 and Eq. 2-3 to show

K_f+U_f=K_i+U_i\qquad\text{(2-11)}

0+mgh=\frac12mv_1^2+0\qquad\text{(2-12)}

gh=\frac12v_1^2\qquad\text{(2-13)}

=\frac12\left(\frac{v_0}2\right)^2\qquad\text{(2-14)}

=\frac12\frac{2gy}4\qquad\text{(2-15)}

h=y/4\qquad\text{(2-16)}

[3 pt]

(b) (12 points)

Define $v_3$ to be the speed of the large block at the top of the loop. For the large block to just barely make it over the top of the loop requires the net force to be equal to the force of gravity:

F_{\mathrm{net}}=3mg=3m\frac{v_3^2}L\qquad\text{(2-17)}

[3 pt]

The last expression comes from the condition that the tangential acceleration of the large block is zero at the top of the loop.

The speed of the large block at the top of the loop is related to the speed at the bottom by conservation of energy (Eq. 2-1).

K_f+U_f=K_i+U_i\qquad\text{(2-18)}

\frac123mv_3^2+(3m)g(2L)=\frac123mv_2^2+0\qquad\text{(2-19)}

v_3^2+4gL=v_2^2\qquad\text{(2-20)}

[3 pt]

Combining the last expression (Eq. 2-20) with the expression for net force (Eq. 2-17) yields

v_3^2+4gL=v_2^2\qquad\text{(2-21)}

gL+4gL=v_2^2\qquad\text{(2-22)}

5gL=v_2^2\qquad\text{(2-23)}

[3 pt]

The combine this result (Eq. 2-23) with the results of part (a), Eq. 2-10. Consequently

v_2=\frac{v_0}2\qquad\text{(2-24)}

v_2^2=\frac{v_0^2}4\qquad\text{(2-25)}

5gL=\frac{2gy}4\qquad\text{(2-26)}

L=y/10\qquad\text{(2-27)}

[3 pt]

Оригинал в Архива: Quarterfinal 2006.pdf · официални решения: Quarterfinal 2006 Solutions.pdf