# 2009年泛珠三角及中华名校物理奥林匹克邀请赛Solution

Pan Pearl River Delta Physics Olympiad 2009

Part-I 第一卷

Q1 题 1 (8 points 8 分) Solution 解: For resistance R, only half of the area is conducting, as the other half is blocked by medium-2. Let the voltage between the plates be V, then the electric field 先求电阻 R.介质-2 不导电,所以只有一半的导电板导电.令两板之间的电压为 V,则 电场为 E = V/d, (1 point 1 分) The current density is 电流密度为 J = σ 1E , (1 point 1 分) The current is 电流为 I = 1 2 a2 V Ja = σ 1 . (1 point 1 分) 2 2 d
V 2d = . (1 point 1 分) I σ 1a 2

So the resistance is 因此电阻为 R =

For capacitance, it can be treated as two capacitors in parallel. The capacitance on the right is

C1 =

ε 0ε1a 2
2d

. (1 point 1 分)

For the left half, let the electric displacement be D2 which is the same throughout the region.

The total free charge on the left half is

Q= a2 D2 . (1 point 1 分) 2

The total voltage between the two plates is

V= 1 1 d 1 1 E1d + E2 d = D2 ( + ) . (1 point 1 分) 2 2 2ε 0 ε1 ε 2

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Pan Pearl River Delta Physics Olympiad 2009

So 因此 C2 =

ε 0 a 2 ε1ε 2 , d ε1 + ε 2

C = C1 + C2 =

ε 0 a 2 ε1

ε1ε 2 + . (1 point 1 分) d 2 ε1 + ε 2

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Pan Pearl River Delta Physics Olympiad 2009

Q2 题 2 (10 points 10 分) (a)

solution 解:

Let the mass of the blackhole be M, then 令黑洞的质量为 M,则 GM v 2 = , (2 points 2 分) r2 r So 因此

GM = v , (1 point 1 分) r

(b)

1 n = . (1 point 1 分) 2 Let the areal mass density be σ, then

Gπσ r 2 v 2 = , (2 points 2 分) r2 r So 因此 Gπσ r = v , (1 point 1 分) 1 . (1 point 1 分) 2 Anything reasonable is fine. It DOES NOT have to be dark matter. n= (c)

Q3 题 3 (10 points 10 分)

Solution 解:

Method 1: Using conservation of energy

Equations 方程:

y2

Mx1 = mx 2 , y 2 = ( x 2 x1 ) tan θ = (1 +
(1 point 1 分)

m ) x 2 tan θ M

k m x2

M

All coordinates are in the rest frame. 所有坐标取

x1

θ

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Pan Pearl River Delta Physics Olympiad 2009

The total kinetic energy of the system is 总动能为
T= 1 1 2 1 2 & & & Mx12 + mx2 + my2 2 2 2 (1 point 1 分) (1 point 1 分)

=

1 m2 2 1 2 1 m & & &2 M 2 x2 + mx2 + m tan 2 θ x2 (1 + ) 2 2 M 2 2 M

The total potential energy 总势能为:

V=

1 2 k[( x 2 x1 ) 2 + y 2 ] mgy 2 (1 point 1 分) 2 1 m m m 2 2 = k[(1 + ) 2 x 2 + (1 + ) 2 x 2 tan 2 θ ] mg (1 + ) x 2 tan θ 2 M M M 1 2 m 1 m = kx 2 (1 + ) 2 mg (1 + ) x 2 tan θ (1 point 1 分) 2 2 M M cos θ

Using 利用 ax 2 bx = a ( x

b 2 b2 ) , we get 得 2a 4a

V=

1 m 1 mMg sin 2θ 2 (mg sin θ )2 k (1 + ) 2 ( x2 ) (1 point 1 分) 2 M cos 2 θ k ( M + m) 2k

Make a transfer of coordinate 取坐标变换

x ≡ x2
We

mMg sin 2θ , (1 point 1 分) k ( M + m)
an expression for the total energy that is of the form of

reach

& T + V = ax 2 + bx 2 + c which must be constant by energy conservation. Let x = A cos(ωt + φ ) , we get T + V = aA2 cos 2 (ωt + φ ) + bA2ω 2 sin 2 (ωt + φ ) + c = A2 (a bω 2 ) cos 2 (ωt + φ ) + bA2ω 2 + c . For the total energy to be constant the cos 2 (ωt + φ ) term must be zero all the time, which leads to ω 2 = a / b . & 我们得到总能量的表达式为 T + V = ax 2 + bx 2 + c .因能量守恒总能量应为常数.令 x = A cos(ωt + φ ) ,得 T + V = aA2 cos 2 (ωt + φ ) + bA2ω 2 sin 2 (ωt + φ ) + c = A2 (a bω 2 ) cos 2 (ωt + φ ) + bA2ω 2 + c

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Pan Pearl River Delta Physics Olympiad 2009

Therefore, the oscillation frequency ω in this case is 由上式得系统的频率ω为 m 2 m ) / cos 2 θ 1+ k k M +m M M ω = = = 2 m m m + m + m(1 + ) tan 2 θ m cos 2 θ + (1 + ) sin 2 θ m M + m sin θ 2 M M M
2

k (1 +

. (1 point 1

1 1 For θ = 0 , we have 当 θ = 0 ,得 ω 2 = k + m M For θ = 90° we have 当 θ = 90°,得 ω 2 =
k . m

(1 point 1 分),

(1 point 1 分)

Method 2: Analytical Mechanics 方法-2:分析力学
y1 k m x1 x2 m

Fspring=-kx1/cosθ

N
mg

M

θ

Force figure 力图 (2 points 2 分) Equations 方程
m x1 =
kx1 cos θ N sin θ + m x 2 , cos θ

x1 is in the frame on the slope. x1 是相对与斜面的横坐标. (1 point 1 分) y1 = x1 tan θ

m x1 tan θ = N cos θ mg kx1 tan θ M x 2 = ( N sin θ + kx1 ) Process 解方程过程 Step 1: Eliminate x 2 步骤-1:消去 x 2
m x 1 = kx1 N sin θ

(1 point 1 分)

(1 point 1 分)

m ( N sin θ + kx1 ) M
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Pan Pearl River Delta Physics Olympiad 2009

m x 1 tan θ = N cos θ mg kx1 tan θ Step 2: Eliminate N
m x 1 + (1 +

m m )kx1 = N sin θ (1 + ) M M

m x 1 tan θ + mg + kx1 tan θ = N cos θ Which means 整理后得 m cos θ x 1 + cos θ (1 +
m m )kx1 = sin θ (1 + )[m tan θ x 1 + mg + kx1 tan θ ] (2 points 2 分) M M

By assuming x1 = Aeiωt , the oscillation frequency ω is obtained

ω2 =

k (1 + m

m ) M

sin 2 θ k M +m cos θ = 2 m sin θ m M + m sin 2 θ cos θ + (1 + ) M cos θ cos θ +

. (1 point 1 分)

1 1 For θ = 0 , we have 当 θ = 0 ,得 ω 2 = k + m M For θ = 90° we have 当 θ = 90°,得 ω 2 = k . m

(1 point 1 分),

(1 point 1 分)

Q4 题 4 (10 points 10 分)

Solution 解:

(a)

Let the magnetic field be B ( z , t ) = B0 e

r

r

% i ( kz ωt )

r %r . The k-vector of the wave is k = kz0 .

r r B Using the equation × E = , t r r r i ( kz ωt ) r r % B r % 令磁场的表达式为 B ( z , t ) = B0 e .k-矢量为 k = kz0 .利用方程 × E =
t

r r r k × E ( z, t ) we get 得 B ( z , t ) = , (1 point 1 分)

ω

=

% k

ω

% k r % % r r E0 ( z0 × x0 )ei ( kz ωt ) = E0 y0 ei ( kz ωt )

ω

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Pan Pearl River Delta Physics Olympiad 2009

% r k r So 因此 B0 = E0 y0 . (1 point 1 分)

ω

(b)

% ω ε +iσ Note that 利用 k = c 2

0 ≡ k R + ik I . εε 0

r r r r r 1 1 So 得 < S >= < ( E × B ) >= Re( E × B* ) (1 point 1 分) 0 20

=

E02 r 1 k R 2 r 2 k I z 1 r z0 Re(k R ik I )e2 kI z = E0 z0 e E02 z0 e 2 kI z = 2 0ω 2 0 ω 2 0 c

(1 point 1 分)

(c)

q=

d < S > k I k R 2 2 kI z = E e 0ω 0 dz

0 εε 0 2 2 k z σ 2 2 k z σ c = = E0 e E0 e . (2 points 2 分) 0ω 2 2 ω ε
I I

(d)

r r r r r r 1 σ σ Joule Heat 焦尔热 = < J E >= Re( J E * ) = Re( E E * ) = E02 e 2 kI z 2 2 2

(2 points 2

(e)

The energy loss of EM wave is equal to the Joule Heat.

Q5 (12 points)

(a) First law 热力学第一定律: dU = PdV + δ Q . The equation of adiabatic processes is 绝热过程的方程式为

δ Q = dU + PdV = d ( 3PV ) + PdV = 4 PdV + 3VdP = 0 (1 point 1 分)
PV 4/3 = Constant PV 4/3 = 常数 (1 point 1 分)

(b) In the Carnot cycle 在卡诺循环过程中:

( P1 ,V1 )

Isothermal

} →

( P1 ,V2 )

} →

( P2 ,V3 )

Isothermal

} →

} P2 , V4 ) → (

( P1 ,V1 ) .

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Pan Pearl River Delta Physics Olympiad 2009

The heat supplied to the gas during the first isothermal process is 第一个等温过程中吸收的热量为
Q1 = 3PV2 3PV1 + ∫ PdV = 4 P (V2 V1 ) . (1 point 1 分) 1 1 1 1
V2 V1

(c) Similarly, the heat supplied to the gas during the second isothermal process is

Q2 = 4 P2 (V4 V3 ) .
(d) From (a) we have 由(a)得: By definition 由定义

(1 point 1 分) (1 point 1 分)

PV24/3 = PV34/3 1 2 PV44/3 = PV14/3 2 1

P (V V ) P1/ 4 P 3/ 4 (V3 V4 ) P1/ 4 T1 Q = 1 = 1 2 1 = 1 2 = 11/ 4 . (2 points 2 分) T2 Q2 P2 (V4 V3 ) P2 (V4 V3 ) P2
Therefore, one may define the absolute temperature by T = AP1/ 4 , where A is an arbitrary constant. Since T = 1 when P = 1, T = P1/ 4 .

(1 point 1 分) (e) The internal energy is then 内能为 U = 3T 4V .
(1 point 1 分) (1 point 1 分)

U 3 Hence the heat capacity is 因此热容量为 CV = = 12T V . T V
(f) The entropy is 熵为 S = ∫ CV
0 T

T dT = 12V ∫ T 2 dT = 4T 3V = 4 P 3/4V . (2 points 2 分) 0 T

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Pan Pearl River Delta Physics Olympiad 2009

Part-II 第二卷

Q1 (16 points) (a)

Consider a thin layer of gas of unit area and thickness dr. The pressure at r should be GMmn larger than the pressure at r + dr in order to balance the gravity . So we r2 GM s mn dP = , where n is the molecular number density of the gas. have dr r2 考虑离太阳 r 处一厚度为 dr 的单位面积气体.在 r 处的气压应比在 r + dr 处的大 一点,从而平衡太阳的引力 分子数密度.(1 point 1 分)
GM s mn GMmn dP dr .因此有 = ,其中 n 为气体的 2 r dr r2

We also have the ideal gas law P = nkT0. Replace P with n we get

GM s m dr dn = . n kT0 r 2

GM s m . kT0

GM s m dr dn = .(1 point 1 分) n kT0 r 2

Finally, ρ = ρ 0eα / r , where α = 最后得 ρ = ρ 0eα / r ,其中 α = (b)

GM s m .(1 point 1 分) kT0

When r > ∞ , ρ > ρ 0 instead of zero. That means the gas ball is infinitely large, which is unphysical. 当 r > ∞ , ρ > ρ 0 而不是零.这意味著气体球是无限大的,不符合实际情况.(2 points 2 分)

(c)

The amount of energy per second through any concentric sphere shells should be constant. That is, J 0 = 4π r 2 I . 每秒钟穿过任意一个同心圆壳的能量应为常数.所以 J 0 = 4π r 2 I .(1 point 1 分)
Then 从而得
J0 = I . (2 points 2 分) 4π r 2 J0 dT dT I (r ) = σ , so 因此 = , (1 point 1 分) dr dr 4πσ r 2
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(d)

Pan Pearl River Delta Physics Olympiad 2009

Then 由此得 T =

. 4πσ r The integral constant should be zero, as T should be zero at large distance.

J0

(e)

GM s mn dP = . dr r2 J But now P = nkT (r ) = 0 kn . Replace n by P in the first equation, we 4πσ r
Again 重力和压力平衡, get

4π GM s mσ P dP = , dr kJ 0 r 4πσ r
β

J0

kn .将 P 代入 n,得 4π GM s mσ . kJ 0

4π GM s mσ P dP = (1 point 1 分) dr kJ 0 r

r which leads to P = P0 r0 r 从而得 P = P0 r0
β

, where β =

,其中 β =
β

4π GM s mσ .(1 point 1 分) kJ 0

4πσ mrP0 r ρ= . (1 point 1 分) kJ 0 r0 This time P and ρ go to zero at large r. 现在的 P 和ρ在无穷远处为零. (f) From the surface temperatures of the planets we know today we estimate that r0 is about the radius of the orbit of Mars.

Q2 (16 points) 题 2(16 分) (a)

Solution 解:

The number of electrons crossing the junction per second is I/e.

On average, there are (α 0.5) I / e electrons flip their spins.

The net angular momentum change per second is then (α 0.5) Ih ×2. 4π e

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Pan Pearl River Delta Physics Olympiad 2009

Ih × 2 .(1 point 1 分) 4π e h I. This is equal to the torque so τ = (α 0.5) 2π e h 角动量的净变化等于力矩 τ = (α 0.5) I .(1 point 1 分) 2π e

(b)

The equation of motion is J

d 2θ dθ = η κθ . 2 dt dt

d 2θ dθ 导线扭摆的运动方程为 J 2 = η κθ .(1 point 1 分) dt dt
% % Let θ (t ) = θ 0 eiωt , where ω is complex, % % 令 θ (t ) = θ 0 eiωt ,其中 ω 是复数,we get 得(1 point 1 分)

% % ω 2 iγω ω02 = 0 ,

2 where 其中 ω0 ≡ κ / J , γ ≡ η / J . (1 point 1 分)

% Let ω = ωR + iωI and solving the equation, we get θ = θ 0 e ωI t eiωR t , % 令 ω = ωR + iωI ,幷解上述方程,得 θ = θ 0 e ωI t eiωR t ,(1 point 1 分)
2 Where 其中 ωI = γ / 2 , (1 point 1 分) , ωR = ω0 + γ 2 / 4 . (1 point 1 分)

(c)

This is a forced oscillation with the force given by τ (t ) = (α 0.5)

h I 0eiωt = τ 0 eiωt . 2e

h I 0eiωt = τ 0 eiωt .(1 point 1 分) 2e

d 2θ dθ The equation of motion is 运动方程为 J 2 = η κθ + τ (t ) . (1 point 1 分) dt dt Let 令 θ (t ) = θ 0 eiωt , (1 point 1 分) we get the oscillation amplitude θ0 =

τ0 / J . ω ω 2 + iγω
2 0

τ0 / J .(1 point 1 分) ω ω 2 + iγω
2 0

The speed of the side wing is 边翼的速度为 v(t ) = iω dθ0 eiωt , (1 point 1 分) and the electromotive potential is 电动势为 ξ (t ) = BLv(t ) = iω dBLθ0 eiωt .(1 point 1 分)

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Pan Pearl River Delta Physics Olympiad 2009

Q3 (18 points) (a)

r r r The magnetic dipole experiences a torque τ = m × B0 which is always perpendicular to

r r r r r the S ~ z0 plane. The torque will turn the direction of S so S rotates around B0 at
constant angular speed.

r r r r r r 磁偶极子受到的力矩为 τ = m × B0 ,其方向始终与 S ~ z0 平面垂直.力矩改变 S 的方 r r 向,因此 S 绕著 B0 以匀角速度旋转.(1 point 1 分) r r Let the angle between S and z0 beθ. The torque is mB0 sin θ = SB0 sin θ , while the
change of angular momentum over time δt is δ S = S sin θδφ .

r r 令 S 与 z0 之间的夹角为θ.则力矩的大小为 mB0 sin θ = SB0 sin θ ,而角动量的变化

Since 既然 S sin θδφ = δ S = SB0 sin θδ t (1 point 1 分) We have 我们得 ω0 = (b)

δφ = B0 . (1 point 1 分) δt

r r In the reference frame rotating at angular velocity ω0 z0 = B0 z0 , the spin appears
stationary.

r r 在以角速度 ω0 z0 = B0 z0 旋转的参照系裏,自旋是不动的.(2 points 2 分)
(c) (d)

r r ω The effective B-field is 有效磁场为 Bω = . (2 points 2 分)

r r In the rotating frame of ω1 z0 , B1 also appears static. r r 在以角速度 ω1 z0 旋转的参照系裏, B1 是不动的.(1 point 1 分)
Let it be along the X' axis in the rotating frame, the total B-field is

r ω r r' B = ( B0 1 ) z0' + B1 x0

r r ω r r' 令 B1 在旋转参照系裏沿 X' 方向,则总磁场为 B = ( B0 1 ) z0' + B1 x0 . (2 points 2

12

Pan Pearl River Delta Physics Olympiad 2009

(e)

r' r' In this case, only B1 x0 remains. The spin will rotate around the x0 axis at angular

speed ω1 = B1 ,
r' 这时的磁场只剩下 B1 x0 .自旋绕其以角速度 ω1 = B1 旋转, (2 points 2 分)

and the time to flip the spin is 倒转自旋所需的时间为 t =
(f)

1 2π π = . (2 points 2 分) 2 ω1 B1

In this case the spin will rotate around the total B-field given by (c) at angular

ω speed ω = B0 + B12 . ω 这时的磁场由(c)给出.自旋的角速度为 ω = B0 + B12 (3 points 3 分)
2

2

~~~~~~~~~

End 完 ~~~~~~~~

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