题目内容
物体A和B用轻绳相连挂于轻弹簧下静止不动,如图所示,A质量为m,B质量为M,当A、B间的轻绳突然断开后,物体A第一次上升至某一位置时的速度为V,此时物体B下落速度为u,在这段时间里,弹簧弹力对物体A的冲量为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241214540652368.png)
A.mυ B.mυ–Mu C.mυ+Mu D.mυ+mu
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241214540652368.png)
A.mυ B.mυ–Mu C.mυ+Mu D.mυ+mu
D
对A根据动量定理有:
,绳子断开之后B做自由落体运动,所以可得
,综合可得弹簧弹力对物体A的冲量为:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121454221720.png)
故选D
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121454158730.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121454190460.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121454221720.png)
故选D
![](http://thumb2018.1010pic.com/images/loading.gif)
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