题目内容
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455780296.gif)
(1)A、B两物体滑到水平面时的速度大小
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455796212.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455811217.gif)
(2)从开始释放A、B两物体到B物体追上A物体所用的总时间t。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241154558273247.gif)
(1)
m/s
m/s(2)
s
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455842523.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455858433.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455874244.gif)
(1)设A物体的质量为m,B物体的质量为M。A、B两物体下滑过程机械能守恒:
,
(2分)
解得:
m/s(2分)
m/s(2分)
(2)设A、B两物体下滑的时间分别为
、
。由牛顿第二定律求得A、B两物体下滑的加速度均为:
(2分)
由运动学公式:
(2分)
解得:
s
s (2分)
在水平面上两物体均做匀速运动,所以当B物体追上A物体时,
有
(2分) 解得:
s(2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455889580.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455905635.gif)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455842523.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455858433.gif)
(2)设A、B两物体下滑的时间分别为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455952207.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455967210.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455983450.gif)
由运动学公式:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455998395.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115456014400.gif)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115456045491.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115456061509.gif)
在水平面上两物体均做匀速运动,所以当B物体追上A物体时,
有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115456076606.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115455874244.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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