Съдържание

Условие Решение

USAPhO 2000, Полуфинал — A4

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Всички материали от USAPhO 2000

2000 Semi-Final Exam · 20 т.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
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

A1 – 1/3

A hand-held calculator may be used. Its memory must be cleared of data and programs. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value 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·m $^2$ /kg $^2$
Coulomb's constant	$k=1/4\pi\epsilon_0=9.0\times10^9$ N·m $^2$ /C $^2$
Biot-Savart constant	$k_m=\mu_0/4\pi=10^{-7}$ T·m/A
Speed of light in a vacuum	$c=3.0\times10^8$ m/s
Boltzmann's constant	$k_B=1.38\times10^{-23}$ J/K
Avogadro's number	$N_A=6.02\times10^{23}$ (mol) $^{-1}$
Ideal gas constant	$R=N_Ak_B=8.31$ J/mol·K
Elementary charge	$e=1.6\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.6\times10^{-19}$ J
Planck's constant	$h=6.63\times10^{-34}\text{ J}\cdot\text{s}=4.14\times10^{-15}\text{ eV}\cdot\text{s}$
Electron mass	$m=9.1\times10^{-31}\text{ kg}=0.51\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\simeq1+nx\quad\text{for }\lvert x\rvert\ll1$

Условие

Two movable wedges of mass $M$ meet the horizontal plane smoothly. They are free to slide without friction on the plane. See accompanying diagram. A coin of mass $m$ is released from rest at the height $h$ and slides down the left hand wedge. Assume that the coin always slides without friction. Express your answers to the following in terms of given quantities and known constants.

Coin m at height h on two movable frictionless wedges of mass M.

a. What is the maximum height the coin will rise on the right-hand wedge? [10 т.]

b. For what minimum mass $M$ will the coin again contact the left-hand wedge? [10 т.]

Решение

Since friction is assumed non-existent, all collisions are elastic. Both momentum and mechanical energy are conserved.

Let $v=$ speed of coin after it has left the left-hand wedge

$V_L=$ speed of left-hand wedge after coin has left

$H=$ maximum height the coin reaches on the right-hand wedge

$V=$ speed of right-hand wedge and coin when coin is at its highest point on the right-hand wedge and at rest with respect to it

In all of the following, velocities to the right will be positive and velocities to the left will be negative. Equating the energy and momentum of the initial condition (coin at height h) and final condition (coin has completely left left-hand wedge).

Momentum conservation

0=-MV_L+mv

Solving for $V_L$

V_L=\frac mM v\qquad (A4.1)

Energy conservation

mgh=\tfrac12MV_L^2+\tfrac12mv^2

Substituting (A4.1)

2mgh=M\left(\frac mMv\right)^2+mv^2=m\left(\frac mM+1\right)v^2

Solving for $v^2$

v^2=\frac{2Mgh}{m-M}\qquad (A4.2)

a. (10 т.)

Equating the energy and momentum of the initial condition (coin with velocity v, right-hand wedge at rest) and final condition (coin at rest with respect to right-hand wedge).

Momentum conservation

mv=(m+M)V\qquad (A4.3)

Energy conservation

\tfrac12mv^2=\tfrac12(m+M)V^2+mgH\qquad (A4.4)

Combining (a4.3) and (A4.4)

mv^2=(m+M)\left(\frac m{m+M}v\right)^2+2mgH=\frac{m^2}{m+M}v^2+2mgH

Solving for $H$

H=\frac1{2g}\left(1-\frac m{m+M}\right)v^2=\frac1{2g}\left(\frac M{m+M}\right)v^2

Substituting (A4.2) into this

H=\frac1{2g}\left(\frac M{m+M}\right)\left(\frac{2Mgh}{m+M}\right)=\left(\frac M{m+M}\right)^2h.

b. (10 т.)

Let $v_F=$ The speed of the coin after it has left the right-hand wedge

$V_R=$ The speed of the right hand wedge after the coin has slid off it

For the coin to again contact the left-hand wedge $v_F>V_L$ . Equating the momentum and energy before and after the coin’s collision with the right-hand wedge

Momentum conservation

mv=MV_R-mv_F

V_R=\frac mM(v+v_F)

Energy conservation

\tfrac12mv^2=\tfrac12MV_R^2+\tfrac12mv_F^2

Combining the last two equations and dividing by $m$ and multiplying by 2.

v^2=\left(\frac mM\right)(v+v_F)^2+v_F^2

Mv^2=m(v^2+2vv_F+v_F^2)+Mv_F^2

Or regrouping terms

0=(M+m)v_F^2+(2mv)v_F-(M-m)v^2

Solving this quadratic equation for $v_F$

v_F=\frac{-2mv\pm\sqrt{(2mv)^2+4(M+m)(M-m)v^2}}{2(M+m)}=\frac{-2mv\pm\sqrt{4M^2v^2}}{2(M+m)}=\frac{-mv\pm Mv}{(M+m)}

Only the plus sign gives a positive speed and a velocity in the correct direction.

v_F=\left(\frac{M-m}{M+m}\right)v

For the coin to again contact the left-hand wedge $v_F>V_L$ , or

\left(\frac{M-m}{M+m}\right)v>\frac mMv

Dividing by $v$ and cross multiplying, the condition becomes

(M-m)M>(M+m)m

M^2-2Mm-m^2>0

Using the quadratic formula this becomes

M>\tfrac12\left(2m\pm\sqrt{4m^2+4m^2}\right)

Since M must be positive this becomes $M>m(1+\sqrt2)$

Оригинал в Архива: Semifinal 2000.pdf · официални решения: Semifinal 2000 Solutions.pdf