USAPhO 2009, Полуфинал — Задача 4. Question A4

United States Physics Team 2009 — Semifinal Exam · 25 т.

Important Instructions for the Exam Supervisor

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  • This examination consists of two parts.
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Указания

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  • 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.

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Possibly Useful Information. You may use this sheet for both parts of the exam.
g=9.8 N/kgg = 9.8\ \mathrm{N/kg}G=6.67×10−11 N⋅m2/kg2G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}
k=1/4πϵ0=8.99×109 N⋅m2/C2k = 1/4\pi\epsilon_0 = 8.99 \times 10^{9}\ \mathrm{N \cdot m^2/C^2}km=μ0/4π=10−7 T⋅m/Ak_{\mathrm{m}} = \mu_0/4\pi = 10^{-7}\ \mathrm{T \cdot m/A}
c=3.00×108 m/sc = 3.00 \times 10^{8}\ \mathrm{m/s}kB=1.38×10−23 J/Kk_{\mathrm{B}} = 1.38 \times 10^{-23}\ \mathrm{J/K}
NA=6.02×1023 (mol)−1N_{\mathrm{A}} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}R=NAkB=8.31 J/(mol⋅K)R = N_{\mathrm{A}} k_{\mathrm{B}} = 8.31\ \mathrm{J/(mol \cdot K)}
σ=5.67×10−8 J/(s⋅m2⋅K4)\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s \cdot m^2 \cdot K^4)}e=1.602×10−19 Ce = 1.602 \times 10^{-19}\ \mathrm{C}
1eV=1.602×10−19 J1\mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}h=6.63×10−34 J⋅s=4.14×10−15 eV⋅sh = 6.63 \times 10^{-34}\ \mathrm{J \cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV \cdot s}
me=9.109×10−31 kg=0.511 MeV/c2m_e = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}(1+x)n≈1+nx(1+x)^n \approx 1 + nx for ∣x∣≪1\lvert x \rvert \ll 1
sin⁡θ≈θ−16θ3\sin\theta \approx \theta - \frac{1}{6}\theta^3 for ∣θ∣≪1\lvert\theta\rvert \ll 1cos⁡θ≈1−12θ2\cos\theta \approx 1 - \frac{1}{2}\theta^2 for ∣θ∣≪1\lvert\theta\rvert \ll 1

Условие

A potato gun fires a potato horizontally down a half-open cylinder of cross-sectional area AA. When the gun is fired, the potato slug is at rest, the volume between the end of the cylinder and the potato is V0V_0, and the pressure of the gas in this volume is P0P_0. The atmospheric pressure is PatmP_{\mathrm{atm}}, where P0>PatmP_0 > P_{\mathrm{atm}}. The gas in the cylinder is diatomic; this means that Cv=5R/2C_{\mathrm{v}} = 5R/2 and Cp=7R/2C_{\mathrm{p}} = 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.

A horizontal cylinder (barrel) with a shaded disk at the left marking the closed end and an open ellipse at the right marking the open end; a dark cylindrical slug inside near the left is labelled 'potato'; a double arrow under the barrel spans its whole length and is labelled L, with 'closed end' written at the left and 'open end' at the right.

The parameters P0P_0, PatmP_{\mathrm{atm}}, V0V_0, and AA are fixed, but the overall length LL of the barrel may be varied.

a. What is the maximum kinetic energy Emax⁡E_{\max} with which the potato can exit the barrel? Express your answer in terms of P0P_0, PatmP_{\mathrm{atm}}, and V0V_0.

b. What is the length LL in this case? Express your answer in terms of P0P_0, PatmP_{\mathrm{atm}}, V0V_0, and AA.

Отговори

Решение

Оригинал в Архива: 2009_SemiFinal.pdf · официални решения: USAPhO-2009-Solutions.pdf