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Условие Отговори Решение

USAPhO 2004, Полуфинал — Задача 4. A4

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Всички материали от USAPhO 2004

AAPT UNITED STATES PHYSICS TEAM AIP 2004 — 2004 Semi-Final Exam · 25 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. 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.
Do not discuss the contents of this exam with anyone until after the submission deadline.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_0 = 8.99 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.00 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_\mathrm{A} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_\mathrm{A}k_\mathrm{B} = 8.31\ \mathrm{J/(mol\cdot K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
Elementary charge	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}$
Electron mass	$m = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ , $\cos\theta \approx 1 - \frac{1}{2}\theta^2$

Условие

A ball is thrown so that it hits a stationary ramp and then bounces up the ramp. Immediately before the first collision between the ball and the ramp, the ball has a velocity $v_\mathrm{o}$ that is horizontal as shown in the accompanying diagram. Assume that all collisions are perfectly elastic, that there is no friction between the ramp and the ball, and that air resistance is negligible. At the instant that the ball makes its $N$ +1 $^{\mathrm{st}}$ collision with the ramp, the ball is moving perpendicular to the ramp. Express your answers in terms of $N$ , $v_\mathrm{o}$ , and $g$ .

A right-triangle ramp rising to the right; a horizontal arrow labelled vo points at the lower part of the incline.

a. Find the angle that the ramp makes with the horizontal. [20 т.]

b. What is the distance, measured along the ramp, between the point at which the first collision occurs and the point at which the $N$ +1 $^{\mathrm{st}}$ collision occurs? [5 т.]

Отговори

Решение

A4. Selecting the $y$ -axis perpendicular to the ramp and the $x$ -axis parallel to the ramp in the upward direction, the components of the gravitational acceleration are

$a_x = -g\sin\theta$ and $a_y = -g\cos\theta$ ,

where $\theta$ is the angle the ramp makes with the horizontal. The components of the initial velocity are $v_{0x} = v_0\cos\theta$ and $v_{0y} = -v_0\sin\theta$ .

a. At each collision with the plane, the $x$ -component of velocity does not change while the $y$ -component reverses sign. At the start of the first bounce $v_{0y} = +v_0\sin\theta$ .

The $y$ -displacement is given by

y = y_0 + v_{0y}t + \tfrac{1}{2}a_y t^2 \qquad (\mathrm{A4-1})

Let $y_0 = 0$ at $t = 0$ , the start of the first bounce. Let $t = t_1$ , the time when the ball returns to the ramp $y = 0$ , at the end of the first bounce. Substituting these values into (A4 – 1)

0 = v_0\sin\theta t_1 - \tfrac{1}{2}g\cos\theta t_1^2.

Solving for $t_1$ and eliminating the initial time $t_1 = 0$ ,

t_1 = \frac{2v_0\sin\theta}{g\cos\theta} = \frac{2v_0}{g}\tan\theta.

The velocity at the end of the first bounce, as the ball is about to impact the ramp again, is

v_y = v_{0y} + a_y t = v_0\sin\theta - g\cos\theta t = v_0\sin\theta - g\cos\theta\frac{2v_0\sin\theta}{g\cos\theta} = -v_0\sin\theta.

Upon hitting the ramp the velocity reverses to become $v_{0y} = +v_0\sin\theta$ .

Each bounce has the same $v_{0y}$ and $a_y$ , so each bounce takes the same amount of time $t_1$ .

Therefore the time for $N$ bounces is

t_N = Nt = N\frac{2v_0\sin\theta}{g\cos\theta} = \frac{2Nv_0}{g}\tan\theta. \qquad (\mathrm{A4-2})

There is no impulse in the $x$ -direction, so the $x$ -equations hold continuously. At the end of the Nth bounce, the ball’s velocity is perpendicular to the ramp, $v_x = 0$ . Substituting this into the equation for the $x$ -component of velocity,

v_x = v_{0x} + a_x t

0 = v_0\cos\theta - g\sin\theta t_N = v_0\cos\theta - g\sin\theta\left(N\frac{2v_0\sin\theta}{g\cos\theta}\right).

Dividing by $v_0\cos\theta$ ,

0 = 1 - \frac{2N\sin^2\theta}{\cos^2\theta} = 1 - 2N\tan^2\theta.

Solving for $\tan\theta$

\tan\theta = \frac{1}{\sqrt{2N}} \qquad (\mathrm{A4-3})

b. The x-displacement is given by $x = v_{0x}t + \frac{1}{2}a_x t^2$ .

The maximum displacement occurs at the end of the Nth bounce, time $t_N$ . Combining (A4 – 2) and (A4 – 3)

t_N = \frac{2Nv_0}{g}\tan\theta = \frac{2Nv_0}{g}\frac{1}{\sqrt{2N}} = \frac{v_0}{g}\sqrt{2N}.

Substituting this into the x-equation,

x = (v_0\cos\theta)\left(\frac{v_0}{g}\sqrt{2N}\right) + \tfrac{1}{2}(-g\sin\theta)\left(\frac{v_0}{g}\sqrt{2N}\right)^2 = \frac{v_0^2}{g}\cos\theta\sqrt{2N} - \frac{v_0^2}{2g}2N\sin\theta

Using the triangle to the right to determine $\cos\theta$ and $\sin\theta$

\cos\theta = \sqrt{\frac{2N}{2n + 1}} \qquad \sin\theta = \frac{1}{\sqrt{2N - 1}}

x = \frac{v_0^2}{g}\sqrt{\frac{2N}{2N + 1}}\sqrt{2N} - \frac{v_0^2}{2g}2N\frac{1}{\sqrt{2N + 1}} = \frac{v_0^2}{g}\frac{N}{\sqrt{2N + 1}}

Right triangle with angle θ at the left vertex: hypotenuse labelled √(2N+1), horizontal leg labelled √(2N), vertical leg labelled 1.

Оригинал в Архива: Semifinal 2004.pdf · официални решения: Semifinal 2004 Solutions.pdf