题目内容
如图所示,内壁光滑半径为R的圆形轨道,固定在竖直平面内.质量为m1的小球静止在轨道最低点,另一质量为m2的小球(两小球均可视为质点)从内壁上与圆心O等高的位置由静止释放,运动到最低点时与m1发生碰撞并粘在一起.求
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250046182383091.jpg)
⑴小球m2刚要与m1发生碰撞时的速度大小;
⑵碰撞后,m1?m2能沿内壁运动所能达到的最大高度(相对碰撞点).
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250046182383091.jpg)
⑴小球m2刚要与m1发生碰撞时的速度大小;
⑵碰撞后,m1?m2能沿内壁运动所能达到的最大高度(相对碰撞点).
(1)
;(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004618269815.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004618253508.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004618269815.png)
试题分析:(1)设小球m2刚要与m1发生碰撞时的速度大小为v0,由机械能守恒定律可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004618300868.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004618316621.png)
(2)设两球碰撞后,m1m2两球粘在一起的速度为v,由动量守恒定律可得:
m2vo=(m1+m2)v ③
两球碰撞后上升的最大高度为h,由机械能守恒定律可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250046183311094.png)
由②③④可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004618347905.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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