XX省浏阳一中、株洲二中等湘东六校届高三物理联考试题

如图所示，甲图为沿x轴传播的一列简谐横波在t=0时刻的波动图象，乙图为参与波动质点P的振动图象，则下列判断正确的是()（填正确答案标号。选对1个得2分，选对2个得3分，选对3个得4分。每选错1个扣2分，最低得分为0分）A．该波的传播速率为4m/sB．该波的传播方向沿x轴正方向C．经过0.5s，质点P沿波的传播方向向前传播2mD．该波在传播过程中若遇到4m的障碍物，能发生明显衍射现象E．经过0.5s时间，质点P的位移为零，路程为0.4m（2）（8分）如图，一个三棱镜的截面为等腰直角形ABC，∠A为直角，直角边长为L.一细束光线沿此截面所在平面且平行于BC边的方向射到AB边上的某点M，光进入棱镜后直接射到BC边上。已知棱镜材料的折射率为，光在真空中速度为c．①作出完整的光路图（含出射光线，并在图上标出角度大小）②计算光从进入棱镜，到第一次射出棱镜所需的时间。

湖南省湘东六校2018年下期高三联考物理参考答案一、选择题：题号12345678910答案BDCAACBCDADBC实验题：（1）3.251.79C（1）E、F、G（2），I1：电流表A1的示数，I2：电流表A2的示数计算题：················2分················2分················1分················1分得：F=12000N设触发“紧急制动”时汽车速度大小为v1,离障碍物距离为x2················1分················1分································1分x=x1+x2················2分得：v1=8m/s,················2分x2=4m紧急制动前的时间为t1,紧急制动后的时间为t2················1分················1分················1分················1分················1分················1分14、设小球A、B在电场中分别运动了t1、t2时间后相遇，两小球加速度大小：················1分 竖直向下················1分················1分 竖直向上················1分两小球在电场中做类平抛运动，设相遇点为P，如图所示，由几何关系得：································1分x方向： ················1分················2分y方向： ················1分················2分代入数据解得：t1=2s，v0=30m/s ················1分⑵设相遇点P的坐标为(xP，yP················1分················2分 ················2分 ················1分带入数据得：=－9m ················1分相遇点P的坐标为(60m，－9m) 15、⑴设物块m1下滑到B点时的速度为v0，由机械能守恒可得：················1分················1分 解得：=6m/s ················2分B················2分 解得： 则碰撞过程中损失的机械能为：················1分································1分················2分解得： ⑵物块由B到C满足机械能守恒：·······2分 ·······2分解得： 设物块滑上木板后，当木板速度为v=2m/s时，物块速度为，·······1分·······1分 解得： 设在此过程中物块运动的位移为x1，木板运动的位移为x2，对物块由动能定理得：·······1分 解得： ·······1分·······1分·······1分对木板M： 解得： 此时木板静止，物块m到木板左端的距离为：·······1分 ·······1分·······1分设物块在台阶上运动的最大距离为x4·······1分·······1分 ·······1分解得： 四、选考题： （1）ACE（4分）(2)设初态气体压强为p1，放砂后压强为p2，体积为V2，升温后体积为V3，则有·······1分p1＝p0＋＝1.5p0·······1分·······1分p2＝p0＋＝3p0·······1分·······2分等温过程：由p1V0＝p2V·······2分·······1分得V2＝0.5V·······1分·······1分·······2分等压过程：由eq\f(V2,T0)＝eq\f(V3,2T0)·······1分·······2分得V3＝V0.（1）ADE（4分）①如图（3分）②作出三角形ABC关于BC边对称图形可知光线在棱镜中的传播路径长度等于MN的距离，即：·······2分