题目内容
如图所示,电荷量均为+q、质量分别为m和2m的小球A和B,中间连接质量不计的细绳,在竖直方向的匀强电场中以速度v0匀速上升,某时刻细绳突然断开.求:
(1)电场强度大小及细绳断开后两球A、B的加速度;
(2)当球B速度为零时,球A的速度大小。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241355363612365.jpg)
(1)电场强度大小及细绳断开后两球A、B的加速度;
(2)当球B速度为零时,球A的速度大小。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241355363612365.jpg)
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536470989.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536377551.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536470989.png)
(1)设电场强度为E,把小球A、B看作一个系统,由于绳未断前两球均做匀速运动,
则
,![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536704770.png)
细绳断后,根据牛顿第二定律得
,
方向向上;
,
(负号表示方向向下).
(2)细绳断开前后两绳组成的系统满足合外力为零,所以系统总动量守恒.设B球速度为零时,A球的速度为vA,根据动量守恒定律得![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536470989.png)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536533704.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536704770.png)
细绳断后,根据牛顿第二定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536751749.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536813555.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536891810.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536377551.png)
(2)细绳断开前后两绳组成的系统满足合外力为零,所以系统总动量守恒.设B球速度为零时,A球的速度为vA,根据动量守恒定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135536470989.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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