题目内容
如图所示,在光滑的水平面上静止着一个质量为m2小球2,质量为m1的小球1以一定的初速度v1朝着球2运动,如果两球之间、球与墙之间发生的碰撞均无机械能损失,要使两球还能再碰,则两小球的质量需满足怎样的关系?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250039489081428.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250039489081428.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003948923632.png)
试题分析: 解析:设两球碰后的速度分别为v1′和v2′,由系统动量守恒定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003948939730.png)
由于发生的是弹性碰撞,碰撞前后的总动能不变,得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250039489551026.png)
联立式①、②,可解得:
球1碰后速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003948970886.png)
球2碰后速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003948986830.png)
按照题意,只要碰后球1不反弹,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003949001472.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003949017505.png)
或者球1反弹,但是其碰后速率-v1′小于球2速率v2′,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003948923632.png)
综上,只要
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825003948923632.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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