题目内容
如图甲所示,一个小弹丸水平射入一个原来静止的单摆并留在里面,结果单摆的振动图线如图乙所示.已知摆球质量为小弹丸质量的5倍,试求小弹丸射入摆球的速度是多大?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241141020523418.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241141020523418.jpg)
0.94 m/s
弹丸水平击中摆球的过程动量守恒,从图中可以读出振幅A和周期T,可求振动的最大速度,进而可求弹丸的速度.
设小弹丸的速度为v0,击入摆球后,摆球的速度为vm
最大偏角为θm,摆球质量为m.
根据机械能守恒定律有:
Ekm=Epm
即
mvm2=mgl(1-cosθm)
=2mgl·sin2![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102161379.gif)
≈2mgl(
·
)2
=
×
A2
解之得:vm=
A
又由单摆周期公式有:
T=2π![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102770300.gif)
由以上各式得:vm=
A
从图乙中可以读出A="0.1" m,T="4" s.
则摆球振动的最大速度为
vm=
=
m/s
根据动量守恒定律,有
mv0=(m+5m)vm
所以v0=6×
m/s="0.94" m/s
设小弹丸的速度为v0,击入摆球后,摆球的速度为vm
最大偏角为θm,摆球质量为m.
根据机械能守恒定律有:
Ekm=Epm
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102146225.gif)
=2mgl·sin2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102161379.gif)
≈2mgl(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102146225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102192233.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102146225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102504386.gif)
解之得:vm=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102738292.gif)
又由单摆周期公式有:
T=2π
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102770300.gif)
由以上各式得:vm=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102785357.gif)
从图乙中可以读出A="0.1" m,T="4" s.
则摆球振动的最大速度为
vm=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102801388.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102816374.gif)
根据动量守恒定律,有
mv0=(m+5m)vm
所以v0=6×
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114102816374.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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