第四届泛珠三角物理奥林匹克竞赛

In the process energy and momentum are conserved, so碰撞前后动量和能量守恒，因此00A B Bx B By A A m v m v m v m v ==+ (1 point )222202A A A B Bx B By m v m v m v m v E =+++ (1 point )Where, A v is in the y-direction, we can get 其中A v 是沿Y-方向的速度。

由此得A v =(1 point ) 0A Bx B m v v m =，A By A B m v v m =−.(1 point )Q2: (6 points )From Gauss theorem, the area density of the free charge at the surfaces of the conductor plates is f D σ=. (0.5 point s) 根据高斯定理，导电板上自由电荷面密度与板间电位移的关系为f D σ=。

In the air gap 130/E E D ε==, while in the dielectric 20/E D εε=. (0.5 point s)在空气中130/E E D ε==，在介质内20/E D εε=，电压为1230011(2)(233333f d d d d dD V E E E σεεεε=++=+=+, 由此得0321f V d εεσε=+. (1 point ) The bound charge is 0012(1)3()21b V E E d εεσεε−=−=+m m . (1 point ) 介质上、下界面的束缚电荷面密度为0012(1)3()21b V E E d εεσεε−=−=+m m 。

When the slab is moving with speed v , the electric currents at two sides are b K v σ=±. From Ampere law, 130B B == (1 point )介质板运动时，上、下界面的束缚电流面密度为b K v σ=±。

由安培定理得在空隙间的磁场130B B ==Inside the dielectric slab the magnetic field is 在介质板里的磁场02000(1)321b V B K v v d εεμμσμε−===+. (1 point )When the parallel conductor plates are moving with speed -v , the electric currents at two plates are 03'21f V K v v d εεσε==+m m . From the Ampere law, 1230f B B B v μσ===−. (1 point ) 当导电板运动时，上、下板的面电流密度为03'21f V K v v d εεσε==+m m 。

由安培定理得 1230f B B B v μσ===−The electron spin (angular momentum) is I , and the associated magnetic dipole moment is 2ge M I m =−r r . In the B-field, the torque on the spin is M B ×r r and perpendicular to M r . (1 point ) The procession frequency is then determined by MB t I Δ=Δ, (1 point ) MB t I θΔ=Δ(1 point )MB I I t θωΔ⇒==Δ (1 point ) which leads to 2ge B m ω=. (1 point ) The negative sign in 2ge M I m=−r r ensures that the spin turns in the same direction as the electron trajectory under Lorentz force. 2emv eB eBv R m ω′=⇒= (2 point s) 电子的自旋（角动量）为I ，与其相关的磁偶极子为2ge M I m =−r r . 在磁场里磁偶极子受的力矩为M B ×r r 与M r 垂直。

(1 point ) M r 进动的频率由下式可求：MB t I Δ=Δ, (1 point )MB t I θΔ=Δ(1 point )MB I I tθωΔ⇒==Δ (1 point ) 得2ge B mω=. (1 point ) 空间的圆轨迹为：2emv eB eBv R m ω′=⇒= (2 point s) 2ge M I m=−r r 中的负号保证了自旋的旋转方向与它在空间的圆轨迹一致。

Q4 (12 point s)(a) Consider a thin layer of air at rest, the pressure difference balances the gravity,取一薄层空气，上、下面上的压强差刚好与重力平衡，()()p h h p h m g hdp m g dhρρ+Δ−=−Δ⇒=− (1 point )(b) Put in the ideal gas law p RT ρ=, the differential equation is then代入理想气体方程p RT ρ=，得微分方程dp mg m g p dh RTρ=−=−. (1 point )(c) From (b), 由(b)解得 θΔI Δr θΔI Δr00ln ln ()mgRT dp mg mgdh p p h p h p e p RT RT −=−⇒−=−⇒=. (3 point s) Put in the numbers, 代入数值，10.0299.818.313008.8mg km RT −×==×. So the height is 得高度为8.8ln 28.80.693 6.1km ×=×=. (1 point )(d) With a constant wind with velocity v we replace the pressure equation by theBernoulli’s equation 有风时，微分方程为 22dp mv d mg dh dhρρ+=−. Using the ideal gas law, we obtain 代入p RT ρ=理想气体方程，得2(1)2dp mv mg p dh RTRT +=−. (1 point ) Therefore 解得 220()mg hmv RT p h p e−⎛⎞+⎜⎟⎜⎟⎝⎠=. (2 point )(e) 22500000/(2460606060)0.01 /v v m s g πΩ×≈Ω×=×××××≈<<r r. (1 point ) (f) The decay length in this case is 代入数值2408.313000.50.029 2.510/3.62ln 2()0.693 6.10.24 6.3 0.0299.8mv RT h kmmg+×+×××==×=+=× (2 point s)Q5 (11 point s):(a) When the ball arrives at point A, it begins to drop down. In this process the potential energy transforms into kinetic energy.球到A 点后下落，到B 点时，势能变化为 (1cos )P MgR θΔ=− (1 point )The moment of inertia about the edge is 绕球边的转动惯量为2275II MR MR =+=%. Kinetic energy is 动能为22221117722510K I MR Mv ωω==×=% (0.5 point s)So 因此22277(1cos )1010K Mv MgR Mv θ′==−+ (0.5point s)2210(1cos )7v gR v θ′⇒=−+2210(1cos )7v g R R ωθ⎛⎞′⇒=−+⎜⎟⎝⎠(1point ) (b) The critical condition for the ball to keep contact with point A before it touches point B is that: At the moment it touches point B, the centrifugal force equals the gravity component.要保持与A 点接触，即球以A 点作圆周运动，其向心力全由重力提供。

2cos Mv Mg R θ′= (1 point ) 22max max 10(1cos )cos (17cos 10)77gR gR v gR v θθθ⇒−+=⇒=− (1point )where cos θ=(c) In the process of the ball collide with point B, the angular momentum of the ball around point B is unchanged. Before the collision, the angular momentum of theball around point B is ()22cos 2cos 25I MRvI MR R MR ωωωθωθ⎛⎞′′′′′+=+=+⎜⎟⎝⎠%. After the collision, it is 275IMR ωω′′′′=%. 在与B 点碰撞过程中，球相对于该点的角动量守恒。

碰撞前的角动量为()22cos 2cos 25I MRvI MR R MR ωωωθωθ⎛⎞′′′′′+=+=+⎜⎟⎝⎠%，碰撞后的角动量为275IMR ωω′′′′=%。