题目内容
如图,质量分别为mA和mB的两小球带有同种电荷,电荷量分别为qA和qB,用绝缘细线悬挂在天花板上。平衡时,两小球恰处于同一水平位置,细线与竖直方向间夹角分别为q1与q2(q1>q2)。两小球突然失去各自所带电荷后开始摆动,最大速度分别为vA和vB,最大动能分别为EkA和EkB。则不正确的是
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004060573888.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004060573888.jpg)
A.mA一定小于mB | B.qA一定大于qB | C.vA一定大于vB | D.EkA一定大于EkB |
B
试题分析:分别对A、B进行受力分析,如图所示
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004060883981.jpg)
两球间的库仑斥力是作用力与反作用力总是大小相等,与带电量的大小无关,因此B错误;
对于A球:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406104661.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406135821.png)
对于B球:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406150669.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406166842.png)
联立得:F=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406182987.png)
在两球下摆的过程中根据机械能守恒:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004061971248.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406275963.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004062911248.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406322967.png)
开始A、B两球在同一水平面上,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406338729.png)
由于θ1>θ2可以得出:LA>LB
这样代入后可知:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406353493.png)
A到达最低点的动能:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004063693900.png)
B到达最低点的动能:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250004064004048.png)
由于θ1>θ2可知,
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406416790.png)
又:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406338729.png)
可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000406447969.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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