第二泛珠三角物理奧林匹克競(jìng)賽答案

1、pan pearl river delta physics olympiad 2006q1kinetic energy of meson 介子動(dòng)能. internal energy of meson介子內(nèi)能suggested steps建議步驟:(1)use energy and momentum conservation to obtain the speed of m2 after collision with m1.利用動(dòng)量能量守恒求得m2與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系統(tǒng)的質(zhì)心速度vc，進(jìn)而求得其動(dòng)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. 介子內(nèi)能等于m2的初始動(dòng)能減介子動(dòng)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，當(dāng)滑輪的轉(zhuǎn)動(dòng)

4、慣量最大時(shí)，重物到達(dá)地面的速度最小。因此m2應(yīng)分布在輪邊上。moment of inertia 滑輪的轉(zhuǎn)動(dòng)慣量無滑動(dòng)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壓強(qiáng)正比于深度作用于條面的壓力為 (b) step-1find the shape of the liquid surface 第一步，求液面形狀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在轉(zhuǎn)動(dòng)參照系內(nèi)，慣性力場(chǎng)為，其勢(shì)能為，加上原有的重力勢(shì)能gh，以及液面應(yīng)是個(gè)等勢(shì)面，得常數(shù)。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液面的斜率應(yīng)在垂直方向與重力平衡，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，定常數(shù)cconsider the volume during rotation, 液體的總體積不變。for在 , 對(duì)條面的壓力附加的力為q4 (12 points) a) use the ideal gas law, 利用理想氣體定律b) energy conservation 能量守恒 (1) and (1a)equal pressure leads to 活塞兩邊壓強(qiáng)相等(2)and (2a)heat transfer plus work done due to expansion熱傳導(dǎo)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.整個(gè)過程中兩邊壓強(qiáng)相等，以及系統(tǒng)的總能量不變，因此有，得p(t)不

9、變。因此整個(gè)過程是等壓的，可用等壓熱容來處理右室的吸熱膨脹，即。c) using the results in (b), one gets 將(b)中有關(guān)的量代入，可得and .q5(12 points)(a) the e-fields in medium-1 and -2 are 在界面兩邊的電場(chǎng) , 磁場(chǎng), with 其中界面兩邊的磁場(chǎng)using the boundary conditions 邊界條件, at y = 0, dispersion relation色散關(guān)系, one gets 得, and ,solving the equations, one obtains 解得,.(b

10、) 將n代如（a）， 得r = 1(c) use (a) and find the phase of r with the given n1 and n2. 將n代如（a）， 得不償失 , phase shift 位相差 = 45°.q6 (13 points)part-apv=nrt so the pressure goes to zero壓力趨于零. part-b(a) , where n are integers n為整數(shù)。(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 與垂直。(b) the torque causes to spin within the x-y plane, and the angular speed is given by . 力矩使在x-y平面內(nèi)旋轉(zhuǎn)，角速度為。the angle of the y-axis is 轉(zhuǎn)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.找出要使一自旋從90° - dq 轉(zhuǎn)到90°，需要多少電子飛過，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.照題意放電荷qthen 得(1)(2)zero potential on sphere surface球面零電勢(shì) => (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.解兩方程，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.正確part-b(a) (b) q0 and at a distance h0 on the other side of the plane. 電荷q0 放在距離界面 h0 的另一邊。(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 的合電勢(shì)在球面上為零。利用a-部分的答案，得, (d) , , (e) sum over all charges on both sides of the plane. 將界面兩邊的電荷之間的力相加 (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.可見力正比于，并只和比例k0 =r/h0有關(guān)，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. 因所有的鏡像電荷都在球面內(nèi)，界面兩邊每對(duì)電荷的距離必大于，所以，只需保留到n = 1的項(xiàng)。the total for