6.2 单维维系动力的养护

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  单维中动力的养护

  

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  For this whale to leap out of the water, something underwater must be moving in the opposite direction, and intuition tells us it must be moving with relatively high . The water that moves downward is pushed downward by the whale's tail, and that allows the whale to rise up.
  ::为了让鲸鱼从水中跳出,水下一定有东西朝着相反的方向移动,直觉告诉我们,它一定在以相对高的高度移动。 向下移动的水被鲸尾的尾巴推向下,让鲸鱼起立。

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  Conservation of Momentum in One Dimension
  ::单维中动力的养护

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  When and were introduced, we used an example of a batted ball to discuss the impulse and momentum change that occurred with the ball. At the time, we did not consider what had happened to the bat.  According to Newton’s third law , however, when the bat exerted a force on the ball, the ball also exerted an equal and opposite force on the bat. Since the time of the collision between bat and ball is the same for the bat and for the ball, then we have equal forces (in opposite directions) exerted for equal times on the ball AND the bat. That means that the impulse exerted on the bat is equal and opposite (-Ft) to the impulse on the ball (Ft) and that also means that there was a change in momentum of the bat $\begin{array}{r}[-\mathrm{\Delta }(mv{)}_{\text{BAT}}]\end{array}$ that was equal and opposite to the change in momentum of the ball $\begin{array}{r}[\mathrm{\Delta }(mv{)}_{\text{BALL}}]\end{array}$ .
  ::当球被引入并被引入时,我们用一个球棒的例子来讨论球中发生的动力和动力变化。当时,我们没有考虑蝙蝠的遭遇。但是,根据牛顿的第三个法律,当蝙蝠对球施加了力量时,球对球施加了平等和对立的力量。由于蝙蝠和球的碰撞时间对球和球是一样的,因此我们对球和球的(反方向)施加了相同的力量。这意味着对蝙蝠施加的冲动与球的冲动(Ft)是相等的和对的(-Ft),这也意味着蝙蝠的动力变化与球的动力变化是相等和对立的[(mv)BAT]。

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  The change in momentum of the ball is quite obvious because it changes direction and flies off at greater . However, the change in momentum of the bat is not obvious at all. This occurs primarily because the bat is more massive than the ball. Additionally, the bat is held firmly by the batter, so the batter's mass can be combined with the mass of the bat. Since the bat's mass is so much greater than that of the ball, but they have equal and opposite forces, the bat's final velocity is significantly smaller than that of the ball. 
  ::球的动力变化是显而易见的, 因为它改变方向, 飞得更大。 但是, 蝙蝠的动力变化并不明显。 这主要是因为蝙蝠比球大。 此外, 蝙蝠被击打者牢牢地抓住, 所以击打者的质量可以与球的质量相结合。 因为蝙蝠的质量比球的要大得多, 但是它们具有相同和相反的力量, 蝙蝠的最后速度比球的速度要小得多 。

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  Consider another system: that of two ice skaters. If we have one of the ice skaters exert a force on the other skater, the force is called an internal force because both the object exerting the force and the object receiving the force are inside the system. In a closed system such as this, momentum is always conserved. The total final momentum always equals the total initial momentum in a closed system. Conversely, if we defined a system to contain just one ice skater, putting the other skater outside the system, this is not a closed system. If one skater pushes the other, the force is an external force because the receiver of the force is outside the system. Momentum is not guaranteed to be conserved unless the system is closed.
  ::考虑另一个系统: 两个滑冰者。 如果我们有一支滑冰者在另一支滑冰者身上施压, 部队就被称为内部力量, 因为施压的对象和接收力量的对象都在系统中。 在这样的封闭系统中, 势头总是会得到保护。 最后总势头总是等于封闭系统中的初始总势头。 相反, 如果我们定义一个系统只包含一个滑冰者, 将另一个滑冰者置于系统之外, 这并不是一个封闭的系统。 如果一个滑冰者将另一个系统推向另一个系统, 部队就是一种外部力量, 因为接收力量的人在系统之外。 除非系统关闭, 否则不能保证保持动力。

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  In a closed system, momentum is always conserved. Take another example: if we consider two billiard balls colliding on a billiard table and ignore , we are dealing with a closed system. The momentum of ball  $\begin{array}{r}A\end{array}$ before the collision plus the momentum of ball  $\begin{array}{r}B\end{array}$ before collision will equal the momentum of ball  $\begin{array}{r}A\end{array}$ after collision plus the momentum of ball  $\begin{array}{r}B\end{array}$ after collision. This is called the law of  and is given by the equation 

  $$\begin{array}{r}{p}_{A\text{before}}+p_{B\text{before}}=p_{A\text{after}}+p_{B\text{after}}\end{array}$$

  
  ::在一个封闭的系统中, 势头总是会被保持。 举另一个例子: 如果我们考虑两台台球在台球桌上相撞而忽略, 我们处理的是一个封闭的系统。 A号球在碰撞前的动力加上B号球在碰撞前的动力, 等于A号球在碰撞后的动力加上B号球在碰撞后的动力。 这被称为“ 校正” , 由等式 pA 前加pB 前= pAafter+pBafter 给出。

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  Example 1
  ::例1

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  Ball  $\begin{array}{r}A\end{array}$ has a mass of 2.0 kg and is moving due west with a velocity of 2.0 m/s while ball  $\begin{array}{r}B\end{array}$ has a mass of 4.0 kg and is moving west with a velocity of 1.0 m/s. Ball  $\begin{array}{r}A\end{array}$ overtakes ball  $\begin{array}{r}B\end{array}$ and collides with it from behind. After the collision, ball  $\begin{array}{r}A\end{array}$ is moving westward with a velocity of 1.0 m/s. What is the velocity of ball  $\begin{array}{r}B\end{array}$ after the collision?
  ::B球的重量为2.0公斤,正以2.0米/秒的速度向西移动,而B球的重量为4.0公斤,正以1.0米/秒的速度向西移动。B球的重量为1.0米/秒。B球A飞越B球,并从后面相撞。碰撞后,A球向西移动,速度为1.0米/秒。B球在碰撞后的速度是多少?

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  Because of the law of conservation of momentum , we know that

  $$\begin{array}{r}{p}_{A\text{before}}+p_{B\text{before}}=p_{A\text{after}}+p_{B\text{after}}\end{array}$$

  .
  ::由于保持势头的法律, 我们知道pA前加pBfe前=pAfter+pBafter。

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  $\begin{array}{r}{m}_A{v}_A+m_B{v}_B=m_{A}v{{}^{\mathrm{\prime }}}_A+m_{B}v{{}^{\mathrm{\prime }}}_B\end{array}$
  ::mAvA+mBvB=mAv_A+mBv_B

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  $\begin{array}{r}(2.0\text{kg})(2.0\text{m/s})+(4.0\text{kg})(1.0\text{m/s})=(2.0\text{kg})(1.0\text{m/s})+(4.0\text{kg})(v_B{}^{\mathrm{\prime }}\text{m/s})\end{array}$
  :2.0公斤)(2.0米/秒)+(4.0公斤)(1.0米/秒)=(2.0公斤)(1.0米/秒)+(4.0公斤)(vB_平方米)

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  $\begin{array}{r}4.0\text{kg}\cdot \text{m/s}+4.0\text{kg}\cdot \text{m/s}=2.0\text{kg}\cdot \text{m/s}+4v_B{}^{\mathrm{\prime }}\text{kg}\cdot \text{m/s}\end{array}$
  ::4.0千克/米/秒+4.0千克/米/秒=2.0千克/米/秒+4vB+4千克/秒

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  $\begin{array}{r}4v_B{}^{\mathrm{\prime }}=8.0-2.0=6.0\end{array}$
  ::4vB-8.0-2.0=6.0

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  $\begin{array}{r}{v}_B{}^{\mathrm{\prime }}=1.5\text{m/s}\end{array}$
  ::vB=1.5毫秒/秒

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  After the collision, ball  $\begin{array}{r}B\end{array}$ is moving westward at 1.5 m/s.
  ::碰撞后,B号球向西移动1.5米/秒。

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  Example  2  
  ::例2

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  A railroad car whose mass is 30,000. kg is traveling with a velocity of 2.2 m/s due east and collides with a second railroad car whose mass is also 30,000. kg and is at rest. If the two cars stick together after the collision, what is the velocity of the two cars?
  ::一辆质量为30,000美元的铁路车。 公斤的车向东行驶速度为2.2米/秒,与第二辆铁路车相撞,车体质量为30,000公斤,停在休息处。 如果两辆汽车在碰撞后合在一起,两辆汽车的速度是多少?

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  Note that since the two trains stick together, the final mass is m _A +m _B , and the final velocity for each object is the same. Thus the conservation of momentum equation,  $\begin{array}{r}{m}_A{v}_A+m_B{v}_B=m_{A}v{{}^{\mathrm{\prime }}}_A+m_{B}v{{}^{\mathrm{\prime }}}_B\end{array}$ , can be rewritten  $\begin{array}{r}{m}_A{v}_A+m_B{v}_B=\left(m_A+m_B\right)v{}^{\mathrm{\prime }}\end{array}$
  ::请注意,由于两列列列在一起,最后质量为 mA+mB,而每个天体的最后速度相同。因此,对动量方程式的保存, mAvA+mBvB=mAv_A+mBv_B,可以改写 mAvA+mBvB=(mA+mB)v_B

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  $\begin{array}{r}(30,000.\text{kg})(2.2\text{m/s})+(30,000.\text{kg})(0\text{m/s})=(60,000.\text{kg})(v^{\mathrm{\prime }}\text{m/s})\end{array}$
  :30,000公斤)(2.2米/秒)+(30,000公斤)(0米/秒)=(60,000公斤)(v_平方米)

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  $\begin{array}{r}66000+0=60000v^{\mathrm{\prime }}\end{array}$
  ::66000+0=60000v______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  $\begin{array}{r}{v}^{\mathrm{\prime }}=\frac{66000}{60000}=1.1\text{m/s}\end{array}$
  ::66,000000=1.1毫秒/秒

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  After the collision, the two cars move off together toward the east with a velocity of 1.1 m/s.  
  ::相撞后,两辆汽车一起向东移动,速度为1.1米/秒。

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  Use the simulation below to observe how and momentum are exchanged in a bumper car collision. Begin by adjusting the sliders to make an elastic collision with cars of different masses. Then, see what happens to the in an inelastic collision with the same cars. Comparing these two scenarios should help you to gain a deeper understanding of why we always rely on momentum to analyze collisions.
  ::使用下面的模拟来观察在撞车中如何和动力的交换。 首先调整滑动器以便与不同质量的汽车发生弹性碰撞。 然后,看看与同一汽车发生无弹性碰撞后会发生什么。 比较这两种情景可以帮助您更深入地了解为什么我们总是依赖动力来分析碰撞。

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  Summary
  ::摘要

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  • A closed system is one in which both the object exerting a force and the object receiving the force are inside the system.
    ::封闭式系统是指在系统中同时使用武力的物体和接收武力的物体都处于系统之内的封闭式系统。
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  • In a closed system, momentum is always conserved.
    ::在封闭的系统中,总能保持势头。

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  Review
  ::回顾

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  1. A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest.  With what speed does the goalie slide on the (frictionless) ice?
  ::1. 以55米/秒的速度移动的0.111公斤曲棍球球杆被80公斤的守门员在休息时抓获。

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  2. A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block.  The block and the bullet fly off together at 9.0 m/s.  What was the original velocity of the bullet?
  ::2. A 0.050公斤子弹击中5.0公斤固定木块,并嵌入该块内,该块和子弹在9.0米/秒时一起飞走。 子弹的最初速度是多少?

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  3. A 0.50 kg ball traveling at 6.0 m/s due east collides head on with a 1.00 kg ball traveling in the opposite direction at -12.0 m/s.  After the collision, the 0.50 kg ball moves away at -14 m/s.  Find the velocity of the second ball after the collision.
  ::3. 0.50公斤的球以6.0米/秒向东冲撞头顶,1公斤的球以-12.0米/秒朝相反方向向飞行。碰撞后,0.50公斤的球以-14米/秒移动。在碰撞后找到第二个球的速度。

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  4. Two carts are stationary with a compressed spring between them and held together by a thread. When the thread is cut, the two carts move apart.  After the spring is released, one cart  $\begin{array}{r}m=3.00\text{kg}\end{array}$ has a velocity of 0.82 m/s east. What is the magnitude of the velocity of the second cart  $\begin{array}{r}(m=1.70\text{kg})\end{array}$ after the spring is released?
  ::4. 两辆马车是固定的,中间有一个压缩的弹簧,用一根线将两辆马车固定在一起。当线线被切开时,两辆马车分开。弹簧释放后,一辆马车=3.00公斤的速度为东经0.82米/秒。弹簧释放后,第二辆马车(m=1.7公斤)的速度有多大?

  

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  5. Compared to falling on a tile floor, a glass may not break if it falls onto a carpeted floor.  This is because
  ::5. 与落在瓷砖地板上相比,玻璃如果落在地毯地毯地板上,就不得碎碎。

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  1. less impulse in stopping.
     ::停止的冲动更少
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  2. longer time to stop.
     ::停止的时间更长。
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  3. both of these
     ::两者
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  4. neither of these.
     ::都不是这些。

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  6. A butterfly is hit by a garbage truck on the highway.  The force of the impact is greater on the
  ::6. 高速公路上一辆垃圾车撞到蝴蝶,撞击力更大。

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  1. garbage truck.
     ::垃圾车
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  2. butterfly.
     ::蝴蝶 蝴蝶 蝴蝶 蝴蝶 蝴蝶 蝴蝶
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  3. it is the same for both.
     ::两者都是一样的。

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  7. A rifle recoils from firing a bullet.  The speed of the rifle’s recoil is small compared to the speed of the bullet because
  ::7. 发射子弹的步枪后坐力,与子弹的速度相比,步枪的后背力速度小于子弹的速度,因为:

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  1. the force on the rifle is small.
     ::步枪上的力量很小
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  2. the rifle has a great deal more mass than the bullet.
     ::步枪的重量比子弹大得多
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  3. the momentum of the rifle is unchanged.
     ::步枪的动力没有变化
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  4. the impulse on the rifle is less than the impulse on the bullet.
     ::步枪上的脉冲比子弹上的脉冲还小
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  5. none of these.
     ::没有这些。

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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. What is Newton's Cradle?
     ::牛顿的摇篮是什么?
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  2. How does Newton's Cradle work?
     ::牛顿的摇篮如何运作?
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  3. How does a Newton's Cradle show conservation of momentum?
     ::牛顿的摇篮是如何保持势头的?