题目内容
如图所示,半径为2R的
圆弧轨道AB和半径为R的
圆弧轨道BC相切于B点,两轨道置于竖直平面内,在C点的正上方有一厚度不计的旋转平台,沿平台的一条直径上开有两个小孔P、Q,两孔离轴心等距离,旋转时两孔均能到达C点正上方,平台离C点的高度为R,质量为2m的小球1自A点由静止开始下落,在B点与质量为m的2球作弹性碰撞,碰后2球过C点,且恰能无碰撞穿过小孔P.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242114262591797.png)
(1)两球第一次碰撞后2球的速度大小
(2)欲使2球能从小孔Q落下,则平台的角速度w应满足什么条件?(不计所有阻力)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211424418272.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211424418272.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242114262591797.png)
(1)两球第一次碰撞后2球的速度大小
(2)欲使2球能从小孔Q落下,则平台的角速度w应满足什么条件?(不计所有阻力)
(1)8
/3(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242114280221376.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211427304434.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242114280221376.png)
试题分析:(1)设小球1在B点速度大小为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211429067323.png)
2mg*2R=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211430284339.png)
得v0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211431376541.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211427304434.png)
两球碰撞过程满足动量守恒和能量守恒,设两球碰后速度大小为v1和v2,则:
2m v0=2 m v1+m v2
且
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211430284339.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211430284339.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211430284339.png)
所以v2=4 v0/3=8
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211427304434.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211437585898.png)
(2)对碰后B球上升至P的过程,由能量守恒:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211430284339.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211430284339.png)
vP=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824211441001689.png)
而小球做竖直上抛的时间满足:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242114420771005.png)
又因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242114432791110.png)
所以角速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242114443711784.png)
点评:本题考查了在碰撞过程中的动量守恒定律的应用。应用程度较高,在运用过程中要结合能量综合考虑。
![](http://thumb2018.1010pic.com/images/loading.gif)
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