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’94高考物理试题覆盖面广、难度适中,既着重考查了学生的基础知识,又注意考查了学生的思维能力、分析解决问题能力和实验能力,有部分试题比较新颖、灵活.下面就一些较好的、较典型的试题的解答错误进行举例分析.一、抽象思维能力和灵活运用数学知识的能力不足.题(31)一带电质点,质量为m,电量为q,以平行于ox轴的速度ν从y轴上的a点射入图1中第一象限所示的区域.为了使该质点能从x轴上的b点以垂直于ox轴的速度v射出,可在适当的地方加一个垂直于xy平面、磁感应强度为B的匀强磁场.若此磁场仅分布在一个圆形区域内,试求这圆形磁场区域的最小半径.重力忽略不计.分析与解:这是一道较新颖的试题.考查了抽象思维、分析推理以及运用数学解决物理问题的能力.带电质点要从a点水平射入再从b点竖直向下射出,故带电质点必然是在磁场力的作用下作1/4的匀速圆周运动,为确定质点作1/4圆周运动的圆心.过a作ox轴的平行线与过b作oy轴的平行线相交于C点,截取CM=CN=R(R为质点运动半径),作正方形O’MCN如图2则O’即是质点作1/4圆周运动的圆心.弦MN的中点为所求磁场圆形区域的圆心.由Bqv=mv~2/R和r=MN/2=2～(1/2)可求得磁场区域的半径r=2～(1/2)mv/2Bq.
’94 college entrance examination physics test coverage, moderate difficulty, not only examines the basic knowledge of students, but also pay attention to examine the students’ thinking ability, analytical ability to solve problems and experimental ability, some of the questions more novel and flexible. Here are some of the better The analysis of the answers to the typical, more typical questions is analyzed. First, the ability of abstract thinking and the ability to use mathematical knowledge flexibly. Problem (31) An electrically charged particle, mass m, and charge q, parallel to the velocity of the ox axis ν is injected from the point a on the y axis into the area shown in the first quadrant of Fig. 1. In order for the particle to be emitted from point b on the x-axis at a velocity v perpendicular to the ox axis, one can be added where appropriate. A uniform magnetic field perpendicular to the xy plane and having a magnetic induction intensity of B. If this magnetic field is only distributed in a circular region, try to find the minimum radius of this circular magnetic region. Gravity is ignored. Analysis and Solution: This is a relatively novel The examination questions examine the ability of abstract thinking, analytical reasoning, and the use of mathematics to solve physical problems. The charged particle must be injected horizontally from point a and then vertically from point b. Therefore, the charged particle must be made under the influence of magnetic force. 1/4 uniform Circular motion, to determine the center of the circle for 1/4 circular motion of the particle. Over a parallel line of the ox axis intersects with the parallel line of the oy axis over b at point C. Intercept CM=CN=R (R is the radius of particle motion) , make a square O’MCN as shown in Figure 2 O’ is the center of the circle for the particle for 1/4 circular motion. The middle point of the string MN is the center of the circle of the desired magnetic field. By Bqv=mv~2/R and r= MN/2 = 2 ~ (1/2) can be obtained in the magnetic field radius r = 2 ~ (1/2) mv / 2Bq.