题目内容
(8分)如图所示,位于竖直平面上的
圆弧轨道光滑,半径为R,OB沿竖直方向,上端A距地面高度为H,质量为m的小球从A点由静止释放,到达B点时的速度为
,最后落在地面上C点处,不计空气阻力.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242215300034171.png)
求:(1)小球刚运动到B点时的加速度为多大,对轨道的压力多大.
(2)小球落地点C与B点水平距离为多少.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824221527585288.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824221528802502.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242215300034171.png)
求:(1)小球刚运动到B点时的加速度为多大,对轨道的压力多大.
(2)小球落地点C与B点水平距离为多少.
(1)aB=2g(2分);3 mg(3分);(2)s=2
(3分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824221531142540.png)
试题分析:(1)由于小球到B点时的速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824221528802502.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824221533794499.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824221533794499.png)
对小球受力分析可得,FN-mg=maB=2mg(2分)
故小球运动到B点对轨道的压力为FN ="3" mg(1分)。
(2)小球从B点被水平抛出时,竖直方向为自由落体运动,即:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824221535463741.png)
水平方向为匀速直线运动,即s=vBt (1分)
联立以上两式,解之得:s=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824221531142540.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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