题目内容
一个连同装备共有
kg的宇宙行员,脱离宇宙飞船后,在离飞船L=45m处与飞船处于相对静止状态,他带着一个装有
0.5kg氧气的贮氧筒,贮氧筒有个可以使氧气以v=50m/s的速度喷出的喷嘴.宇航员必须向着与返回飞船相反的方向释放氧气,才能回到飞船上去,同时又必须保留一部分氧气供他在飞向飞船的途中呼吸.飞行员呼吸的耗氧率为
.如果他在开始返回的瞬间释放
的氧气,他能安全回到飞船吗?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443518395.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443534330.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443565489.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443565425.gif)
宇航员可以安全返回飞船.
本题立意在分析解决实际问题.宇航员放出氧气后,由于反冲使自己获得返回飞船的速度.设其反冲速度为
,由动量守恒定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443596494.gif)
因
,故有![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443628414.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443643592.gif)
宇航员返回飞船的时间![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443659472.gif)
在这900s内,宇航员需要呼吸氧气
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443674479.gif)
可以看出:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443690495.gif)
所以,宇航员可以安全返回飞船.
如果宇航员以最短的时间返回飞船,设时间为t,宇航员放出氧气的质量为Δm,则留下呼吸的氧气至少为m-Δm.根据动量守恒定律,宇航员获得的反冲速度:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443706499.gif)
故有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443721447.gif)
而宇航员呼吸氧气应满足:
两式联立,可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443752540.gif)
代入数据解出Δm=0.45kg(另一解Δm=0.05kg舍去)
求出![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443768383.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443581305.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443596494.gif)
因
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443612404.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443628414.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443643592.gif)
宇航员返回飞船的时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443659472.gif)
在这900s内,宇航员需要呼吸氧气
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443674479.gif)
可以看出:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443690495.gif)
所以,宇航员可以安全返回飞船.
如果宇航员以最短的时间返回飞船,设时间为t,宇航员放出氧气的质量为Δm,则留下呼吸的氧气至少为m-Δm.根据动量守恒定律,宇航员获得的反冲速度:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443706499.gif)
故有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443721447.gif)
而宇航员呼吸氧气应满足:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443737441.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443752540.gif)
代入数据解出Δm=0.45kg(另一解Δm=0.05kg舍去)
求出
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115443768383.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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