ANÁLISIS DE RESULTADOS

INSTRUCTIONS

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. 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 Prob. 1 - P. 1/3 • A ruler will be required on this exam.

• 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 questions (or their answers or solutions) on this contest until after March 10, 2009.

1. Below is an image of Fomalhaut b, the first extrasolar planet to be observed directly by visible light, obtained by the Hubble Space Telescope.

Astronomy Picture of the Day 2008 November 14 http://antwrp.gsfc.nasa.gov/apod/ap081114.html Image Credit: NASA, ESA, P. Kalas, J. Graham, E. Chiang, E. Kite (Univ. California, Berkeley), M. Clampin (NASA/Goddard), M. Fitzgerald (Lawrence Livermore NL), K. Stapelfeldt, J. Krist (NASA/JPL)

The scale of the larger diagram is shown on the lower left; the 13” refers to the angle, in arc-second, subtended by 100 AU at the distance of Fomalhaut. The scale of the inset can be determined by size of the small box.

(a) A planet is in a circular orbit of radius R and period T . Derive an expression for the mass of a star in terms of R, T , and the gravitational constant G. You may assume that the mass of the planet is much, much less than the mass of the star.

(b) From the image, estimate the mass of the star Fomalhaut in solar masses. You may assume that the orbit of Fomalhaut b is circular and ignore any errors associated with the fact that the plane of the image is not coincident with the plane of the orbit. Recall that the radius of the Earth’s orbit around the Sun is 1 AU. You do not need to do an error analysis, but the number of significant digits in your answer ought reflect the accuracy of your answer.

Solution

The force between the star of mass M and the planet of mass m is given by F = GM m

R2

Assuming that M  m, then the center of mass of the system is located at the star, and therefore the orbital radius of the planet is effectively R. For a circular orbit,

F = mv 2 R = m (2πR/T )2 R = 4π 2_mR T2. Equating, 4π2R T2 = GM R2 ,

or the more familiar Kepler’s law

R3 T2 = GM 4π2 Rearranging, M = 4π2GR 3 T2

To find the mass of Fomalhaut b, Mb in terms of solar masses Ms, we simply require a ratio:

Mb Ms = Rb 3_/T b2 Re3/Te2 ,

where b refers to the planet around Fomalhaut b, while e refers to Earth. Then Mb Ms = Rb Re 3  Te Tb 2 .

From the diagram we can obtain the ratios. Note that the 13 arcsecond notation is not needed for this computation, it simply is a comparison of the size of the object as seen from Earth. As such, seen from Earth, the orbital dust ring subtends about the same angle as does the planet Jupiter! A quick estimate yields Rb/Re≈ 110, while slightly more work is required for the periods.

The inset box is 10 times the scale of the main image.

The planet moves 1.4 AU during the two year period. Since the orbital circumference is 690 AU, that means it takes 980 years to complete an orbit. Then Tb/Te= 980, and Mb/Ms ≈ 1.4. Allowing

for 10% error in each measurement, the results would be between 0.87 and 2.2.

Finally, after marking the quarter-final answers, it was mildly entertaining to consider the range of values obtained, from a minimum of 10−19(about 1011 kg, or the mass of a pile of rocks some 300 meters tall) to a maximum of 1025 (About 1000 times greater than the mass of the universe!).

2. A ball of mass m is thrown vertically upward with a speed of v0. The ball is subject to an

air resistance force that is proportional to the velocity, F = −kv. The ball rises up to height of h and then returns to the starting point after some total time tf. The acceleration of free

fall is g.

Determine the following:

(a) An expression for the velocity as a function of time in terms of any or all of the constants. (b) Sketch the graph of velocity vs time. On the sketch indicate the time to rise to the highest point, tr, and the total time of flight, tf. You do not need to find either tr or tf

at this point, but your graph should reflect relative positions of both.

(c) Find the time to rise, tr, to the highest point, h, in terms of any or all of the constants.

Solution

Two forces act on the ball: gravity, and air friction. They are not necessarily equal, so the net force, and hence an expression for the acceleration, would be given by

ma = −mg − kv

where up is positive. We can’t simply rearrange this to solve for v, since a is a variable. We can, however, rearrange to prepare for integration, since a = dv/dt:

dv dt = −  g + k mv  or dv g + _mkv
= −dt. Integrating both sides

m k ln  mg/k + v mg/k + v0  = −(t − t0).

Assume t0 is zero. Then

v = (mg/k + v0)e−

k

mt− mg/k

Note that as t → ∞ the ball will approach a constant speed of vt = mg/k. Of course, it will

probably hit the ground first, but if it were thrown up in the air over the edge of an infinitely deep cliff....

The ball will rise until the velocity is zero, or when −vt g ln  vt vt+ v0  = tr.

or, if you prefer,

tr = vt g ln  1 +v0 vt 

To find the height we need to integrate one more time. Since v = dx/dt we can write dx =  (vt+ v0)e −g vtt_{− vt}  dt

which integrates quickly to h = (vt+ v0)  −vt g   e− g vttr _{− 1}  − (vttr)

Gluing together with the expression for tr,

h = vtv0 g − vt2 g ln  1 +v0 vt  .

Showing that it reduces to h = v₀2/2g in the limit as k → 0 is left as an exercise for the reader. 3. A parallel plate capacitor is made of two square parallel plates of area A, and separated by

a distance d  √A. The capacitor is connected to a battery with potential V and allowed to fully charge. The battery is then disconnected. A square metal conducting slab also with area A but thickness d/2 is then fully inserted between the plates, so that it is always parallel to the plates. How much work has been done on the metal slab while it is being inserted?

Solution

The capacitance of a thin rectangular capacitor is given by C = 0A/d

If you forgot this, then derive it. The electric field between the plates is given by E = Q

0Ad

so the potential difference between the plates must be V = Ed = Q

0A

and the capacitance is

C = Q/V = 0A/d

For the original capacitor we’ll call this C0.

The energy stored in a capacitor is U = 1₂QV = 1₂CV02, so once charged, the initial energy is

U0 = 1 2 0A d V 2 0

Note, we are going to need to know the initial charge sooner or later, which is Q0 = C0V0=

0AV0

d

Inserting the conducting plate can be treated two ways: it results in the creation of two capacitors, in series, each with area A but differing separation. The net charge Q0 won’t change, that’s why

For our capacitors, the only thing that changes is d, which we’ll call x1 and x2. Then 1/Ce= x1/0A + x2/0A which gives us Ce = 0A d/2 = 2C0 Since the new total separation x1+ x2 is just d − d/2.

The energy in the system after inserting the plate is U = 1 2Q0V = 1 2Q 2 0/Ce

since V is not constant! Finally, U = 1 2Q 2 0/2C0= 1 2U0

The work done on the plate is the same as the work done on the system, which is given by W = U − U0 or W = −1 4 0A d V 2 0.

This means that the plate is “sucked” into the original capacitor, and we have to fight to hold it out.

4. A certain electric battery is not quite ideal. It can be thought of as a perfect cell with constant output voltage V0 connected in series to a resistance r, but there is no way to remove this

internal resistance from the battery.

V r

R

Battery

0

N parallel bulbs (4 are shown)

The battery is connected to N identical lightbulbs in parallel. The bulbs each have a fixed resistance R, independent of the current through them.

(a) Derive an expression for the total power dissipated by the N bulbs in terms of r, R, and V0.

(b) It is observed that the configuration dissipates more total power through the bulbs when N = 5 than it does for any other value of N . In terms of R, what range of values is possible for r?

Solution

The effective resistance of N identical resistors in parallel is Re= R/N

The total resistance of the circuit is the r + R/N . The current drawn is then

I = V0 r + R/N. The power dissipated by the parallel resistance is given by

P = I2Re, so P = V0 2 (r + R/N )2 R N = N R V02 (N r + R)2

One can simply start with the fact that the maximum power is dissipated when the external resistance is equal to the internal resistance. Or, take the derivative of P with respect to R and set it equal to zero. In either case, you will get

r = Re=

R N

Don’t assume, however, that the answer is then r = R/5. Since we can’t have half a bulb, we don’t actually know that the power is a maximum for the circuit, only that it is the maximum configuration. It could be that a larger power output could occur with a resistance between R/4 and R/5, or maybe between R/5 and R/6.

A rough guess would be to assume

R

5.5 ≤ r ≤ R 4.5

A better estimate would be to assume a quadratic function of 1/N with respect to P in the vicinity of Pmax(think Taylor expansion of P ), and then the important point is the halfway point between

1/N and 1/(N ± 1), or 1/N + 1/(N ± 1) 2 = 1 2N + 1 2(N ± 1) = N ± 1/2 N (N ± 1) So the bounds are

5.5R 30 ≤ r ≤

4.5R 20

The correct approach is to assume the power outputs at N and N ± 1 are equal, and solve for the necessary value of r. Then

or

N3r2+ 2N2(R ± r)r + N (R ± r)2 = N3r2+ 2N2Rr + N R2± N2r2± 2N Rr ± R2 Thankfully, we loose the cubic term

±2N2r2± 2N Rr + N r2 = ±N2r2± 2N Rr ± R2. To save space, flip the ± to the one term that doesn’t have it.

N2r2± N r2_{= R}2

Only positive answers have meaning, so

r = 1

pN (N ± 1)R which give the correct bounds as

R √ 30 ≤ r ≤ R √ 20

United States Physics Team

Semi Final Contest

AAPT

UNITED STATES PHYSICS TEAM

AIP

2009

Semifinal Exam

DO NOT DISTRIBUTE THIS PAGE

Important Instructions for the Exam Supervisor

• This examination consists of two parts.

• Part A has four questions and is allowed 90 minutes. • Part B has two questions and is allowed 90 minutes.

• The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam.

• The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining. • Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B.

• Allow 90 minutes to complete Part B. Do not let students go back to Part A.

• Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-4), and Part B (pages 6-7). Examinees should be provided parts A and B individually, although they may keep the cover sheet.

• The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after March 31, 2009.

• Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.

AAPT

UNITED STATES PHYSICS TEAM

AIP

2009

Semifinal Exam

INSTRUCTIONS

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

• Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time.

• After you have completed Part A you may take a break.

• Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time.

• Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets.

• Start each question on a new sheet of paper. Put your school ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example,

School ID # Doe, Jamie

A1 - 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, PDA’s or cameras 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. • Questions with the same point value are not necessarily of the same difficulty.

• In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after March 31, 2009.

Possibly Useful Information. You may use this sheet for both parts of the exam. g = 9.8 N/kg G = 6.67 × 10−11N · m2/kg2 k = 1/4π0= 8.99 × 109 N · m2/C2 km= µ0/4π = 10−7 T · m/A c = 3.00 × 108 m/s kB = 1.38 × 10−23J/K NA= 6.02 × 1023(mol)−1 R = NAkB= 8.31 J/(mol · K) σ = 5.67 × 10−8J/(s · m2_{· K}4_{) e = 1.602 × 10}−19_C 1eV = 1.602 × 10−19J h = 6.63 × 10−34J · s = 4.14 × 10−15eV · s me= 9.109 × 10−31kg = 0.511 MeV/c2 (1 + x)n≈ 1 + nx for |x|  1

Part A

Question A1

A hollow cylinder has length l, radius r, and thickness d, where l  r  d, and is made of a material with resistivity ρ. A time-varying current I flows through the cylinder in the tangential direction. Assume the current is always uniformly distributed along the length of the cylinder. The cylinder is fixed so that it cannot move; assume that there are no externally generated magnetic fields during the time considered for the problems below.

l

r I

a. What is the magnetic field strength B inside the cylinder in terms of I, the dimensions of the cylinder, and fundamental constants?

b. Relate the emf E developed along the circumference of the cylinder to the rate of change of the current dI_dt, the dimensions of the cylinder, and fundamental constants.

c. Relate E to the current I, the resistivity ρ, and the dimensions of the cylinder. d. The current at t = 0 is I0. What is the current I(t) for t > 0?

Solution

The magnetic field through the inside of the cylinder is given by B = µ0I/l

so the magnetic flux is

ΦB = BA = πµ0r2I/l

The inductance is then

L = ΦB/I = πµ0r2/l

Induced emf as a function of changing current is then E = −LdI dt = − πµ0r2 l dI dt But that induced emf will be driving the current, so

where R is the resistance, given by

R = ρLength Area

Here the legnth is the circumference, 2πr, while the area is the cross sectional area of the conductor, ld. Therefore,

E = Iρ2πr ld Combining the above, we get a differential equation,

Iρ2πr ld = − πµ0r2 l dI dt which can be written more simply as

−αI = dI dt where

α = 2ρ µ0rd

and the solution is then

I(t) = I(0)e−αt

Question A2

A mixture of32P and35S (two beta emitters widely used in biochemical research) is placed next to a detector and allowed to decay, resulting in the data below. The detector has equal sensitivity to the beta particles emitted by each isotope, and both isotopes decay into stable daughters.

You should analyze the data graphically. Error estimates are not required. Day Activity Day Activity Day Activity

0 64557 40 12441 200 1121

5 51714 60 6385 250 673

10 41444 80 3855 300 467

20 27020 100 2734

30 18003 150 1626

a. Determine the half-life of each isotope. 35S has a significantly longer half-life than 32P. b. Determine the ratio of the number of 32P atoms to the number of 35S atoms in the original

sample.

Question A3

Two stars, each of mass M and separated by a distance d, orbit about their center of mass. A planetoid of mass m (m  M ) moves along the axis of this system perpendicular to the orbital plane.

z d

axis perpendicular to plane of orbit

Let Tp be the period of simple harmonic motion for the planetoid for small displacements from

the center of mass along the z-axis, and let Ts be the period of motion for the two stars. Determine

the ratio Tp/Ts.

This problem was adapted from a problem by French in Newtonian Mechanics.

Solution

This problem starts out similar to the quarterfinal exam. The center of mass of the two stars is given by

R = d/2. The force of attraction between the two stars is given by

F = GM2/d2 The orbital period T will be given by the centripetal force law

F = M4π 2_R T2 = G M2 d2 or Ts= 2π r d3 2GM = 4π r R3 GM

The planetoid is a distance z above the plane. The distance to either star is then √R2_{+ z}2_.

Only the part of the force that is perpendicular to the plane survives the vector addition, so the net force is given by

F = 2 GmM R2_{+ z}2

z √

R2_{+ z}2.

This expression is exact. But if z  d, then we can approximate this as F ≈ 2GmM

R3 z

This is a restoring force; the planetoid will execute simple harmonic oscillations with period Tp = 2π

r m k where k is the effective force constant, or

Tp= 2π r R3 GM = 2π r d3 8GM so Tp/Ts= 1/2

Question A4

A potato gun fires a potato horizontally down a half-open cylinder of cross-sectional area A. When the gun is fired, the potato slug is at rest, the volume between the end of the cylinder and the potato is V0, and the pressure of the gas in this volume is P0. The atmospheric pressure is Patm,

where P0 > Patm. The gas in the cylinder is diatomic; this means that Cv= 5R/2 and Cp= 7R/2.

The potato moves down the cylinder quickly enough that no heat is transferred to the gas. Friction between the potato and the barrel is negligible and no gas leaks around the potato.

L

closed end open end potato

The parameters P0, Patm, V0, and A are fixed, but the overall length L of the barrel may be

varied.

a. What is the maximum kinetic energy Emaxwith which the potato can exit the barrel? Express

your answer in terms of P0, Patm, and V0.

b. What is the length L in this case? Express your answer in terms of P0, Patm, V0, and A.

Solution

So long as the pressure inside the cylinder is greater than the external air pressure, the potato will accelerate. Therefore, maximum energy will be transferred to the potato if the cylinder is exactly long enough so that the final pressure inside the volume of the cylinder is Patm.

The energy of an ideal diatomic gas is given by CvnRT

but, by the ideal gas law,

P V = nRT so the energy in the gas is

CvP V

for any volume and pressure. Maximum energy is delivered to the potato when the final pressure is atmospheric, so the work done by the gas on the potato is

Cv(P0V0− PatmVf)

But the potato is moving against air, so the actual energy given to the potato is Emax= Cv(P0V0− PatmVf) − Patm(Vf − V0)

The relationship determining Vf is that of adiabatic expansion,



where γ = Cp/Cv. Things don’t simplify much, unless we use the numerical values of Cv and γ. Then Emax= 5 2P0V0− 7 2PatmVf + PatmV0, or Emax=  5 2P0+ Patm− 7 2Patm 2/7_P 05/7  V0

The length of the tube is simply

L = Vf/A = V0 A  P0 Patm 5/7 .

STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for

Part A, but do not continue to Part B until instructed by your exam

Part B

Question B1

A bowling ball and a golf ball are dropped together onto a flat surface from a height h. The bowling

In document La producción de textos ficcionales y el desarrollo de la creatividad en estudiantes universitarios (página 56-63)