题目内容
(12分)如图所示,质量为M、内有半径R的半圆形轨道的槽体放在光滑的平台上,左端紧靠一台阶,质量为m的小物体从A点由静止释放,若槽内光滑。 求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250003587723678.jpg)
(1)小物体滑到圆弧最低点时的速度大小v
(2)小物体滑到圆弧最低点时,槽体对其支持力N的大小
(3)小物体上升的最大高度h
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250003587723678.jpg)
(1)小物体滑到圆弧最低点时的速度大小v
(2)小物体滑到圆弧最低点时,槽体对其支持力N的大小
(3)小物体上升的最大高度h
(1)
;(2)
;(3)
R
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358787508.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358803512.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358818623.png)
试题分析:(1)设小物体由A落至圆弧最低点时的速度为v,取圆弧最低点为势能零点,
由机械能守恒定律得:mgR=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358850338.png)
得v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358787508.png)
(2)在最低点对小球受力分析,由
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358896859.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250003589121015.png)
(3)小物体向上运动的过程中,m与M组成的系统在水平方向的动量守恒:
设小球滑至最高点时m与M的共同速度为v′
所以 mv=(M+m)v′ 2分
解得:v′=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358943901.png)
此过程中系统机械能守恒,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358850338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358850338.png)
解得m上升的最大高度h=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000358818623.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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