题目内容
两球A、B在光滑水平面上沿同一直线,同一方向运动,mA="1" kg,mB="2" kg,vA="6" m/s,vB="2" m/s。当A追上B并发生碰撞后,两球A、B速度的可能值是( )
A.
,
B.
, ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211715847611.png)
C.
,
D.
, ![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211719123750.png)
A.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211713398614.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211714209697.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211715020620.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211715847611.png)
C.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211716518601.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211717438608.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211717875732.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211719123750.png)
BD
试题分析:考虑实际情况,碰撞后A球速度不大于B球的速度,因而A错误,BCD满足;两球碰撞过程,系统不受外力,故碰撞过程系统总动量守恒,ABD满足;根据能量守恒定律,碰撞后的系统总动能应该小于或等于碰撞前的系统总动能,碰撞前总动能为22J,B选项碰撞后总动能为18J,D答案
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211719497494.png)
点评:本题碰撞过程中动量守恒,同时要遵循能量守恒定律,不忘联系实际情况。
![](http://thumb2018.1010pic.com/images/loading.gif)
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