题目内容
(18分)如图,竖直固定轨道abcd段光滑,长为L=1.0m的平台de段粗糙,abc段是以O为圆心的圆弧.小球A和B紧靠一起静止于e处,B的质量是A的4倍.两小球在内力作用下突然分离,A分离后向左始终沿轨道运动, 与de段的动摩擦因数μ=0.2,到b点时轨道对A的支持力等于A的重力的
, B分离后平抛落到f点,f到平台边缘的水平距离S= 0.4m,平台高h=0.8m,g取10m/s2,求:
(1)(8分)AB分离时B的速度大小vB;
(2)(5分)A到达d点时的速度大小vd;
(3)(5分)圆弧 abc的半径R.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241222403035387.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122240287222.gif)
(1)(8分)AB分离时B的速度大小vB;
(2)(5分)A到达d点时的速度大小vd;
(3)(5分)圆弧 abc的半径R.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241222403035387.jpg)
(1)vB="1" m/s
(2)vd= 2
m/s
(3)R=0.5m
(2)vd= 2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122240350225.gif)
(3)R=0.5m
(1)解: (1)B分离后做平抛运动,由平抛运动规律可知:
h=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122240381225.gif)
(2)AB分离时,由动量守恒定律得:
mAve=mBvB ……...2分) A球由 e到d根据动能定理得:
-μmAgl=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122240381225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122240381225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122240350225.gif)
(3)A球由d到b根据机械能守恒定律得:
mAgR+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122240381225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122240381225.gif)
mAg-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122240287222.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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