题目内容
(9分)如下图所示,在竖直平面内固定着半径为R的半圆形轨道,小球B静止在轨道的最低点,小球A从轨道右端正上方3.5R处由静止自由落下,沿圆弧切线进入轨道后,与小球B发生弹性碰撞。碰撞后B球上升的最高点C,圆心O与C的连线与竖直方向的夹角为60°。若两球均可视为质点,不计一切摩擦,求A、B两球的质量之比
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250010496142207.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001049598493.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250010496142207.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001049630609.png)
试题分析:小球A从高处静止下落至轨道的最低点,由机械能守恒定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250010496451029.png)
小球A与小球B发生弹性碰撞,由动量守恒定律和能量守恒定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001049661712.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001049676994.png)
B球上升到最高点C,由机械能守恒定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250010496921214.png)
联立解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001049630609.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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