6.4 弹性和弹性碰撞

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  电磁和无弹性碰撞

  

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  This device is known as Newton’s cradle. As the balls collide with each other, nearly all the and is conserved. If one ball swings down, exactly one ball will swing up; if three balls swing down, exactly three will swing back up. The collisions between the balls are very nearly elastic. 
  ::这个装置被称为牛顿的摇篮。 当球相互碰撞时,几乎所有的球都会被保存下来。 如果一个球跌倒,一个球就会被击倒;如果三个球跌倒,一个球就会被击倒;如果三个球跌倒,一个球就会被击倒。 球之间的碰撞几乎是弹性很大的。

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  Elastic and Inelastic Collisions
  ::电磁和无弹性碰撞

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  For all collisions in a closed system , momentum is conserved. In some collisions in a closed system, kinetic energy is conserved. When both momentum and kinetic energy are conserved, the collision is called an elastic collision .  Most collisions are inelastic because some amount of kinetic energy is converted to , usually by raising one of the objects higher (increasing gravitation PE) or by flexing the object. Any denting or other changing of shape by one of the objects will also be accompanied by a loss of kinetic energy. The only commonly seen elastic collisions are those between billiard balls or ball bearings, because these balls do not compress. And, of course, collisions between molecules are elastic if no damage is done to the molecules.
  ::对于封闭系统中的所有碰撞,都保持了动力力。在封闭系统中的某些碰撞中,运动力节能。当动力能和动能节能时,碰撞被称为弹性碰撞。大多数碰撞都是无弹性的,因为有一定数量的动能被转换成,通常是通过提高一个更高的物体(引力增加)或移动物体。如果一个物体的变形或其他形状变化也会伴随动能的丧失。唯一常见的弹性碰撞是球球球或球轴之间的碰撞,因为这些球不会压缩。当然,分子之间的碰撞是弹性的,如果分子没有受损的话。

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  Much more common are . These collisions occur whenever kinetic energy is not conserved, primarily when an object's height is increased after the collision or when one of the objects is compressed.
  ::更常见的是。这些碰撞发生在动能得不到节能时,主要是在碰撞后物体的高度升高时,或当一个物体被压缩时。

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  Example  
  ::示例示例示例示例

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  A 12.0 kg toy train car moving at 2.40 m/s on a straight, level train track, collides head-on with a second train car whose mass is 36.0 kg and was at rest on the track. If the collision is perfectly elastic and all motion is frictionless, calculate the velocities of the two cars after the collision.
  ::12.0公斤的玩具列车在直轨、一级列车轨道上以2.40米/秒的速度行驶,与质量为36.0公斤、休息在轨道上的第二辆列车正面碰撞。如果碰撞完全弹性强,而且所有运动无摩擦,则计算碰撞后两辆车的速度。

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  Since the collision is elastic, both momentum and KE are conserved. We use the and conservation of KE equations.
  ::由于碰撞具有弹性,动力和KE都得到了保护。我们使用并保存了KE方程式。

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  Conservation of momentum: $\begin{array}{r}{m}_1{v}_1+m_2{v}_2=m_{1}v{{}_1}^{\mathrm{\prime }}+m_{2}v{{}_2}^{\mathrm{\prime }}\end{array}$
  ::保持动力: m1v1+m2v2=m1v1=m2v2=m2v2}m2v2

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  Conservation of KE: $\begin{array}{r}\frac{1}{2}{m}_{1}v{{}_1}^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}{m}_1{v{{}_1}^{\mathrm{\prime }}}^2+\frac{1}{2}{m}_2{v{{}_2}^{\mathrm{\prime }}}^2\end{array}$
  ::保护 KE: 12m1v12+12m2v22=12m1v1_2+12m2v2

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  Since $\begin{array}{r}{m}_1,m_2,v_1\end{array}$ , and  $\begin{array}{r}{v}_2\end{array}$ are known, only  $\begin{array}{r}v{{}_1}^{\mathrm{\prime }}\end{array}$ and  $\begin{array}{r}v{{}_2}^{\mathrm{\prime }}\end{array}$ are unknown. When the known values are plugged into these two equations, we will have two equations with two unknowns. Such systems can be solved with algebra.
  ::由于已知 m1,m2,v1 和 v2, 只有 v1 和 v2 未知。 当已知值被插入到这两个方程式中时, 我们将会有两个未知方程式。 这些系统可以用代数解析 。

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  $\begin{array}{r}(12.0kg)(2.40m/s)+(36.0kg)(0m/s)=(12.0kg)(v{{}_1}^{\mathrm{\prime }})+(36.0kg)(v{{}_2}^{\mathrm{\prime }})\end{array}$
  :12.0千克)(2.40米/秒)+(36.0千克)(0米/秒)=(12.0千克)(v1})+(36.0千克)(v2})

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  $\begin{array}{r}28.8=12.0v{{}_1}^{\mathrm{\prime }}+36.0v{{}_2}^{\mathrm{\prime }}\end{array}$
  ::28.8=12.0 v136.0 v2

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  Solving this equation for  $\begin{array}{r}v{{}_1}^{\mathrm{\prime }}\end{array}$ yields $\begin{array}{r}v{{}_1}^{\mathrm{\prime }}=2.4-3v{{}_2}^{\mathrm{\prime }}\end{array}$
  ::解决 v1 收益率 v1\\\\ 2.4-3 v2 的这个方程式

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  $\begin{array}{r}\frac{1}{2}(12.0)(2.40{)}^2+\frac{1}{2}(36.0)(0{)}^2=\frac{1}{2}(12.0)(v{{}_1}^{\mathrm{\prime }}{)}^2+\frac{1}{2}(36.0)(v{{}_2}^{\mathrm{\prime }}{)}^2\end{array}$
  ::12(12.0)(2.40)2+12(36.0)(0)2=12(12.0)(v1)2+12(36.0)(v2})2

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  $\begin{array}{r}69.1=12.0{v{{}_1}^{\mathrm{\prime }}}^2+36.0{v{{}_2}^{\mathrm{\prime }}}^2\end{array}$
  ::69.1=12.0 v1_2+36.0 v2_2

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  $\begin{array}{r}5.76={v{{}_1}^{\mathrm{\prime }}}^2+3{v{{}_2}^{\mathrm{\prime }}}^2\end{array}$
  ::5.76=v1=2+3 v2=2

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  Substituting the equation for  $\begin{array}{r}v{{}_1}^{\mathrm{\prime }}\end{array}$ into this equation yields
  ::将 v1 的方程转换为此方程产量

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  $\begin{array}{r}5.76=(2.4-3v{{}_2}^{\mathrm{\prime }}{)}^2+3{v{{}_2}^{\mathrm{\prime }}}^2\end{array}$
  ::5.76=(2.4-3 v2_)2+3 v2_2

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  $\begin{array}{r}5.76=5.76-14.4v{{}_2}^{\mathrm{\prime }}+9{v{{}_2}^{\mathrm{\prime }}}^2+3{v{{}_2}^{\mathrm{\prime }}}^2\end{array}$
  ::5.76=5.76-14.4 v29 v2_2+3 v2_2

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  $\begin{array}{r}12{v{{}_2}^{\mathrm{\prime }}}^2-14.4v{{}_2}^{\mathrm{\prime }}=0\end{array}$
  ::12 v2_2_14.4 v2_0

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  $\begin{array}{r}12v{{}_2}^{\mathrm{\prime }}=14.4\end{array}$
  ::12 v214.4

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  $\begin{array}{r}v{{}_2}^{\mathrm{\prime }}=1.2m/s\end{array}$
  ::1.2毫秒/秒

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  Substituting this result back into  $\begin{array}{r}v{{}_1}^{\mathrm{\prime }}=2.4-3v{{}_2}^{\mathrm{\prime }}\end{array}$ , we get $\begin{array}{r}v{{}_1}^{\mathrm{\prime }}=-1.2m/s\end{array}$ .
  ::将这一结果改为 v12.4-3 v2,我们得到v11.2 m/s。

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  So, the heavier car is moving in the original direction at 1.2 m/s and the lighter car is moving backward at 1.2 m/s.
  ::因此,较重的汽车正向原来的方向移动1.2米/秒,较轻的汽车向后移动1.2米/秒。

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  Use the simulation below to verify that momentum is always conserved in a closed system. Create an elastic collision by setting the slider to “bouncy” and observe if both the momentum and kinetic energy are conserved by analyzing the graphs. Then, create an inelastic collision by setting the slider “locking” and see what happens to the exchange of momentum and in the collision.
  ::使用下面的模拟来验证动力总是在封闭系统中保持的。 通过将滑动器设为“自动 ” , 并观察是否通过分析图表来保护动力和动能,从而制造弹性碰撞。 然后,通过设置滑动器“锁住 ” , 并观察动力和动能的交换和碰撞情况,从而造成无弹性碰撞。

   

   

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  Further Reading
  ::继续阅读

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  • Elastic Collisions
    ::电磁碰撞
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  • Energy- Problems
    ::能源问题

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  Summary
  ::摘要

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  • Elastic collisions are those in which both momentum and kinetic energy are conserved.
    ::电磁碰撞是节能动力和动能的碰撞。
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  • Inelastic collisions are those in which either momentum or kinetic energy is not conserved.
    ::无弹性碰撞是指动力或动能得不到保护的碰撞。

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  Review
  ::回顾

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  1. A 4.00 kg metal cart is sitting at rest on a frictionless ice surface.  Another metal cart whose mass is 1.00 kg is fired at the cart and strikes it in a one-dimensional elastic collision.  If the original velocity of the second cart was 2.00 m/s, what are the velocities of the two carts after the collision?
     ::另一辆质量为1.00公斤的金属车向该车发射,并在单维弹性碰撞中撞击。如果第二辆汽车的原速度是2.00米/秒,那么两辆汽车在碰撞后的速度如何?
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  2. Identify the following collisions as most likely elastic or most likely inelastic.
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     1. A ball of modeling clay dropped on the floor.
        ::一球模拟粘土掉在地上
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     2. A fender-bender automobile collision.
        ::摔跤手撞车
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     3. A golf ball landing on the green.
        ::高尔夫球降落在绿地上
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     4. Two billiard balls colliding on a billiard table.
        ::两台台球在台球桌上碰撞
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     5. A collision between two ball bearings.
        ::两颗球轴承的碰撞
     
     ::确定以下碰撞为最有可能有弹性或最有可能无弹性的碰撞。 模拟粘土球落在地板上。 挡风玻璃和大腿的汽车碰撞。 高尔夫球落在绿色上。 两个台球在台球桌上碰撞。 两个球轴承之间的碰撞。

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  Explore More
  ::探索更多

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  Use this resource to answer the questions that follow.
  ::使用此资源回答下面的问题 。

   

   

   

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  1. Explain what happened in the first demonstration on elastic collisions.
     ::解释一下在第一次弹性碰撞示范中发生了什么
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  2. Explain what happened in the second demonstration on inelastic collisions.
     ::解释一下在第二次无弹性碰撞示范中发生了什么
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  3. Assuming the first carts started at the same speed in both demonstrations, explain using momentum why the inelastic collision ended slower than the elastic collision.
     ::假设第一车在两次演示中以同样的速度开动,则用动力解释无弹性碰撞的结束比弹性碰撞慢的原因。

   

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  Resource
  ::资源资源资源资源资源资源资源资源资源资源资源资源

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  What is bouncing? Why do some rubber balls bounce more than others? What does this have to do with energy conversion? Find out in the video below!
  ::什么是弹跳?为什么一些橡胶球比其他橡胶球弹得更多?这与能源转换有什么关系?请在下面的视频中查一查!