题目内容
装甲车和战舰采用多层钢板比采用同样质量的单层钢板更能抵御穿甲弹的射击.通过对以下简化模型的计算可以粗略说明其原因.
质量为2m、厚度为2d的钢板静止在水平光滑桌面上。质量为m的子弹以某一速度垂直射向该钢板,刚好能将钢板射穿.现把钢板分成厚度均为d、质量均为m的相同两块,间隔一段距离平行放置,如图所示.若子弹以相同的速度垂直射向第一块钢板,穿出后再射向第二块钢板,求子弹射入第二块钢板的深度.设子弹在钢板中受到的阻力为恒力,且两块钢板不会发生碰撞。不计重力影响。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241358301141787.png)
质量为2m、厚度为2d的钢板静止在水平光滑桌面上。质量为m的子弹以某一速度垂直射向该钢板,刚好能将钢板射穿.现把钢板分成厚度均为d、质量均为m的相同两块,间隔一段距离平行放置,如图所示.若子弹以相同的速度垂直射向第一块钢板,穿出后再射向第二块钢板,求子弹射入第二块钢板的深度.设子弹在钢板中受到的阻力为恒力,且两块钢板不会发生碰撞。不计重力影响。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241358301141787.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830161739.png)
设子弹初速度为v0,射入厚度为2d的钢板后,
由动量守恒得:mv0 =(2m+m)V(2分)
此过程中动能损失为:ΔE损=f·2d=
mv
-
×3mV2 (2分)
解得ΔE=
mv![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830270293.png)
分成两块钢板后,设子弹穿过第一块钢板时两者的速度分别为v1和V1:mv1+mV1=mv0(2分)
因为子弹在射穿第一块钢板的动能损失为ΔE损1=f·d=
mv
(1分),
由能量守恒得:
mv
+
mV
=
mv
-ΔE损1(2分)
且考虑到v1必须大于V1,
解得:v1=
v0
设子弹射入第二块钢板并留在其中后两者的共同速度为V2,
由动量守恒得:2mV2=mv1(1分)
损失的动能为:ΔE′=
mv
-
×2mV
(2分)
联立解得:ΔE′=
×
mv
因为ΔE′=f·x(1分),
可解得射入第二钢板的深度x为:
(2分)
子弹打木块系统能量损失完全转化为了热量,相互作用力乘以相对位移为产生的热量,以系统为研究对象由能量守恒列式求解
由动量守恒得:mv0 =(2m+m)V(2分)
此过程中动能损失为:ΔE损=f·2d=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830192338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830270293.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830192338.png)
解得ΔE=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830364327.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830270293.png)
分成两块钢板后,设子弹穿过第一块钢板时两者的速度分别为v1和V1:mv1+mV1=mv0(2分)
因为子弹在射穿第一块钢板的动能损失为ΔE损1=f·d=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830457338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830270293.png)
由能量守恒得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830192338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830660281.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830192338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830660281.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830192338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830270293.png)
且考虑到v1必须大于V1,
解得:v1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135831097696.png)
设子弹射入第二块钢板并留在其中后两者的共同速度为V2,
由动量守恒得:2mV2=mv1(1分)
损失的动能为:ΔE′=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830192338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830660281.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830192338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135831331244.png)
联立解得:ΔE′=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135831409694.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830457338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830270293.png)
因为ΔE′=f·x(1分),
可解得射入第二钢板的深度x为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135830161739.png)
子弹打木块系统能量损失完全转化为了热量,相互作用力乘以相对位移为产生的热量,以系统为研究对象由能量守恒列式求解
![](http://thumb2018.1010pic.com/images/loading.gif)
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