题目内容
(16分)如图1所示,一质量为m的滑块(可视为质点)沿某斜面顶端A由静止滑下,已知滑块与斜面间的动摩擦因数
和滑块到斜面顶端的距离
的关系如图2所示。斜面倾角为37°,长为L。有一半径
的光滑竖直半圆轨道刚好与斜面底端B相接,且直径BC与水平面垂直,假设滑块经过B点时没有能量损失。当滑块运动到斜面底端B又与质量为m的静止小球(可视为质点)发生弹性碰撞(已知:
,
)。求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/2014082500110407510705.png)
(1)滑块滑至斜面底端B时的速度大小;
(2)在B点小球与滑块碰撞后小球的速度大小;
(3)滑块滑至光滑竖直半圆轨道的最高点C时对轨道的压力。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001103997301.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104013266.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104028509.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104028651.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104060676.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/2014082500110407510705.png)
(1)滑块滑至斜面底端B时的速度大小;
(2)在B点小球与滑块碰撞后小球的速度大小;
(3)滑块滑至光滑竖直半圆轨道的最高点C时对轨道的压力。
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104106572.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104075737.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104106399.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104106572.png)
试题分析:(1)滑块由顶端滑至底端,由动能定理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250011041381243.png)
由图2的物理意义得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250011041381519.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104075737.png)
(2)滑块与小球发生弹性碰撞,设碰后速度分别为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104169298.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104184330.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104200639.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104387928.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104106399.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104668830.png)
说明:学生未列双守恒方程,直接利用速度交换的结论,只扣1分。
(3)滑块从B到C,由动能定理有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250011046841036.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104699756.png)
对滑块根据牛顿第二定律有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104715903.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104730558.png)
由牛顿第三定律得滑块在C点时对轨道的压力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001104762630.png)
说明:学生未利用牛顿第三定律求压力,只扣1分
![](http://thumb2018.1010pic.com/images/loading.gif)
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