题目内容
如图3所示,一质量为m的小球,在B点从静止开始沿半球形容器内壁无摩擦地滑下,B点与容器底部A点的高度差为h.容器质量为M,内壁半径为R,求:
(1)当容器固定在水平桌面上,小球滑至底部A时,容器内壁对小球的作用力大小.
(2)当容器放置在光滑的水平桌面上,小球滑至底部A时,小球相对容器的速度大小?容器此时对小球的作用力大小.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241154440804695.gif)
(1)当容器固定在水平桌面上,小球滑至底部A时,容器内壁对小球的作用力大小.
(2)当容器放置在光滑的水平桌面上,小球滑至底部A时,小球相对容器的速度大小?容器此时对小球的作用力大小.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241154440804695.gif)
(1)T=mg+m
=mg+m
=mg(1+
)
(2)v′=v1-v2=
T′=mg+m
=mg[1+
]
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444096260.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444111412.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444127376.gif)
(2)v′=v1-v2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444142674.gif)
|
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444158665.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444174627.gif)
(1)m下滑只有重力做功,故机械能守恒,即有
mgh=
mv2,v2=2gh ①
底部A是圆周上的一点,由牛顿第二定律,有:T-mg=m
T=mg+m
=mg+m
=mg(1+
)
(2)容器放置在水平桌面上,则m与M组成的系统在水平方向不受外力,故系统在水平方向上动量守恒;又因m与M无摩擦,故m与M的总机械能也守恒.令m滑到底部时,m的速度为v1,M的速度为v2.
由动量守恒定律得:0=mv1+Mv2 ①
由机械能守恒定律得:mgh=
mv12+
Mv22 ②
联立①②两式解得:v1=
,v2=-![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444298387.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444283671.gif)
小球相对容器的速度大小v′,v′=v1-v2=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444142674.gif)
由牛顿第二定律得:T′-mg=m![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444345267.gif)
T′=mg+m
=mg[1+
]
mgh=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444189225.gif)
底部A是圆周上的一点,由牛顿第二定律,有:T-mg=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444096260.gif)
T=mg+m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444096260.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444111412.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444127376.gif)
(2)容器放置在水平桌面上,则m与M组成的系统在水平方向不受外力,故系统在水平方向上动量守恒;又因m与M无摩擦,故m与M的总机械能也守恒.令m滑到底部时,m的速度为v1,M的速度为v2.
由动量守恒定律得:0=mv1+Mv2 ①
由机械能守恒定律得:mgh=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444189225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444189225.gif)
联立①②两式解得:v1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444283671.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444298387.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444283671.gif)
小球相对容器的速度大小v′,v′=v1-v2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444142674.gif)
由牛顿第二定律得:T′-mg=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444345267.gif)
T′=mg+m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444158665.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115444174627.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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