题目内容
(9分)如图所示,ABC为一固定在竖直平面内的光滑轨道,BC段水平,AB段与BC段平滑连接.质量为m的小球从高为h处由静止开始沿轨道下滑,与静止在轨道BC段上质量为km的小球发生碰撞,碰撞前后两小球的运动方向处于同一水平线上。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242338237133925.jpg)
(1)若两小球碰撞后粘连在一起,求碰后它们的共同速度;
(2)若两小球在碰撞过程中无机械能损失,为使两小球能发生第二次碰撞,求k应满足的条件。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242338237133925.jpg)
(1)若两小球碰撞后粘连在一起,求碰后它们的共同速度;
(2)若两小球在碰撞过程中无机械能损失,为使两小球能发生第二次碰撞,求k应满足的条件。
(1)v=
(2)k>3
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233824556640.png)
试题分析:(1)小球从圆弧面滑下的过程由机械能守恒定律得:mgh=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233826116342.png)
由动量守恒定律得:mv0=(m+km)v,解得v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233824556640.png)
(2)若两小球在碰撞过程中无机械能损失,则动量和机械能守恒,得:
mv0=mv1+kmv2,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233826116342.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233826116342.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233826116342.png)
解得:v1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233832793407.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233834025444.png)
须满足:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233835445395.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824233834025444.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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