题目内容
“∟”形轻杆两边互相垂直、长度均为l,可绕过O点的水平轴在竖直平面内自由转动,两端各固定一个金属小球A、B,其中A球质量为m,带负电,电量为q(q>0),B球开始不带电,质量未知。现将“∟”形杆从OB位于水平位置由静止释放:(sin37°=0.6,cos37°=0.8)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242339455981018.png)
(1)当OB杆转过37°时,两球的速度达到最大,则B球的质量为多少?
(2)若在空间加一竖直向下的匀强电场,OB杆从原来位置开始释放能转过的最大角度为127°,则该电场的电场强度大小为多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242339455981018.png)
(1)当OB杆转过37°时,两球的速度达到最大,则B球的质量为多少?
(2)若在空间加一竖直向下的匀强电场,OB杆从原来位置开始释放能转过的最大角度为127°,则该电场的电场强度大小为多少?
(1)
;(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233947719638.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233946783622.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233947719638.png)
试题分析: (1) 由机械能守恒可知,动能最大时,重力势能最小,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242339487022334.png)
由数学规律可知
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242339497011451.png)
其中
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233951151773.png)
由题意可知,当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233951963455.png)
即:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233952914489.png)
可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233953367760.png)
(2)到达最大角度时,动能为零,由功能关系得:
mAgl(1+sin37°)- mBglcos37°=Eql(1+sin37°) (3分)
可解得:E =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233947719638.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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