第二泛珠三角物理奥林匹克竞赛答案

1、pan pearl river delta physics olympiad 2006q1kinetic energy of meson 介子动能. internal energy of meson介子内能suggested steps建议步骤:(1)use energy and momentum conservation to obtain the speed of m2 after collision with m1.利用动量能量守恒求得m2与m1碰撞后的速度。 (2) find the center-of-mass speed vc of m2-m3 system (m3 is stil

2、l at rest right after collision). the kinetic energy ek of the m2+m3 system is then (m2+m3)/2 times the square of vc. 由(1)求得m2-m3系统的质心速度vc，进而求得其动能。(3)the 'internal energy' ei can then be found using energy conservation, counting both the meson and the electron after collision, or just counti

3、ng the energy m2 has after collision. 介子内能等于m2的初始动能减介子动能。q2when the moment of inertia of the pulley is the largest, which means putting mass along the edge. then use energy conservation that the initial potential energy is equal to the kinetic energy of mass m3 and that of the pulley. 由能量守恒可知，当滑轮的转动

4、惯量最大时，重物到达地面的速度最小。因此m2应分布在轮边上。moment of inertia 滑轮的转动惯量无滑动能量守恒答案q3(10 points)(a) pressure is proportional to depth, and the total force is found by integrating the entire depth. the force acting on the strip is压强正比于深度作用于条面的压力为 (b) step-1find the shape of the liquid surface 第一步，求液面形状method-1: in the

5、rotating reference frame the inertia force is , and the associated potential is . the liquid surface is an equal-potential surface, and the gravitation potential is gh. so the shape is a rotating parabola determined by constant.方法-1在转动参照系内，惯性力场为，其势能为，加上原有的重力势能gh，以及液面应是个等势面，得常数。method-2: the slope of

6、 the surface should be such that normal force should balance the gravity in the vertical direction, and give the concentric force for in the horizontal direction. that leads to the same answer as method-1.方法-2液面的斜率应在垂直方向与重力平衡，在水平方向提供所需的向心力。method-3方法-3 where其中 where其中where is a constant.step-2determ

7、ine c第二步，定常数cconsider the volume during rotation, 液体的总体积不变。for在 , 对条面的压力附加的力为q4 (12 points) a) use the ideal gas law, 利用理想气体定律b) energy conservation 能量守恒 (1) and (1a)equal pressure leads to 活塞两边压强相等(2)and (2a)heat transfer plus work done due to expansion热传导及能量守恒 (3) finally we have (4)from eqs. (1),

8、 (2) and (4) we get (5) also from eq. (1a) (6)put eqs. (5) and (6) into eq. (3) where 其中, and .alternative method:另一方法throughout the process the pressures in both chambers are the same, so, and the pressure is a constant. then.using (6) above we get the same answers.整个过程中两边压强相等，以及系统的总能量不变，因此有，得p(t)不

9、变。因此整个过程是等压的，可用等压热容来处理右室的吸热膨胀，即。c) using the results in (b), one gets 将(b)中有关的量代入，可得and .q5(12 points)(a) the e-fields in medium-1 and -2 are 在界面两边的电场 , 磁场, with 其中界面两边的磁场using the boundary conditions 边界条件, at y = 0, dispersion relation色散关系, one gets 得, and ,solving the equations, one obtains 解得,.(b

10、) 将n代如（a）， 得r = 1(c) use (a) and find the phase of r with the given n1 and n2. 将n代如（a）， 得不偿失 , phase shift 位相差 = 45°.q6 (13 points)part-apv=nrt so the pressure goes to zero压力趋于零. part-b(a) , where n are integers n为整数。(b) . the force is given by 力等于. the system energy decreases with increasing d

11、 so the force is pushing outwards. 力的方向向外。(c) outside we have 板外面, so因此 f = 0 (d) . q7 (15 points)(a) the torque is 力矩 which is perpendicular to 与垂直。(b) the torque causes to spin within the x-y plane, and the angular speed is given by . 力矩使在x-y平面内旋转，角速度为。the angle of the y-axis is 转角 (c) find how ma

12、ny electrons should pass to cause a particular spin to rotation from 90° - dq to 90°and from q find position of the spin.找出要使一自旋从90° - dq 转到90°，需要多少电子飞过，得所以(d) .q8 (22 points)part-aplace a point charge q at distance x from the center, which is at a distance r from a point on the

13、sphere surface. the distance of this point to charge q is l.照题意放电荷qthen 得(1)(2)zero potential on sphere surface球面零电势 => (3) putting eq. (1) and (2) into (3) leads to (4)equation (4) must be true for all angle , so式-4要在所有成立，因此(5) and(6)solving eq. (5) and (6), we get two sets of solutions.解两方程，得so

14、lution-1解-1: x = d and q = q. not the right one because q ends up outside the sphere.不用solution-2解-2: q = qr/d, and x = r2/d < r. correct.正确part-b(a) (b) q0 and at a distance h0 on the other side of the plane. 电荷q0 放在距离界面 h0 的另一边。(c) the contribution of q1 and q2 is to make the sphere surface zer

15、o potential. the answer in part-a can be used here. q1 和 q2 的合电势在球面上为零。利用a-部分的答案，得, (d) , , (e) sum over all charges on both sides of the plane. 将界面两边的电荷之间的力相加 (f) using (a) and (d), define 利用前面的答案，令，得, using (e) . the force is proportional to , and depend only on the ratio of k0 =r/h0. so the answe

16、r is 4.4 x 10-12 n.可见力正比于，并只和比例k0 =r/h0有关，因此的力 = 4.4 x 10-12 n。(g) since all the image charges above the surface must be inside the spheres, the distance between any charge outside the sample and those inside should be large than . then . so, and only the n = 1 terms should be kept. 因所有的镜像电荷都在球面内，界面两边每对电荷的距离必大于，所以，只需保留到n = 1的项。the total for