题目内容
(18分)如图所示,一质量
的足够长木板B静止于光滑水平面上,B的右边放有竖直挡板,B的右端距挡板
.现有一小物体A(可视为质点)质量
,以初速度
从B的左端水平滑上B.已知A与B间的动摩擦因数
,B与竖直挡板的碰撞时间极短,且碰撞时无机械能损失.求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250023248747559.png)
(1)B与竖直挡板相碰前的速度是多少?
(2)若题干中的S可以任意改变(S不能为零)大小,要使B第一次碰墙后,AB系统动量为零,S的大小是多少?
(3)若要求B与墙碰撞两次,B的右端距挡板S应该满足什么条件?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324796620.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324811503.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324827530.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324842623.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324858475.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250023248747559.png)
(1)B与竖直挡板相碰前的速度是多少?
(2)若题干中的S可以任意改变(S不能为零)大小,要使B第一次碰墙后,AB系统动量为零,S的大小是多少?
(3)若要求B与墙碰撞两次,B的右端距挡板S应该满足什么条件?
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324936589.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324889605.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324920568.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324936589.png)
试题分析:(1)设A和B达到共同速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324952298.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324983283.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324998791.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324889605.png)
对B由动能定理有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250023250301036.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325061507.png)
由于
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325076563.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324889605.png)
(2)设B与墙壁第一次碰前A、B的速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325123371.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325201370.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325217809.png)
对B由动能定理有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250023252481066.png)
B与墙碰撞时无机械能损失,则B以速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325201370.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325279721.png)
联立解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324920568.png)
(3)设B与墙壁第一次碰前A、B的速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325342400.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325357390.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325451846.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250023254661058.png)
若要求B与墙碰撞两次则碰后系统的总动量仍向右,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002325482772.png)
可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002324936589.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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