动量守恒

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Conservation of Momentum

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09:46 min
April 30, 2023

Overview

资料来源:尼古拉斯 · 蒂蒙斯Asantha 库雷博士,物理系 & 天文,物理科学学院,加利福尼亚大学,加利福尼亚州欧文市

本试验主要目的是动量的测试守恒的概念。通过设置与非常小的摩擦表面,可以研究了运动物体之间的碰撞,包括其初始的和最终的动量。

动量守恒定律是物理学中最重要的法律之一。当一些东西守恒的物理学中时,初始价值等于最终值。对于动量,这意味着系统总最初的势头将等于总的最终动力。牛顿第二定律物体上的力将等于中对象的动量随时间的变化。这一事实,结合思想动量守恒的是经典力学的运作的基础,是解决问题的有力工具。

Principles

动量Equation 1定义为物体时间为其速度的质量Equation 2:

Equation 3.(方程 1)

此外可以定义对象 (牛顿第二定律) 力推动:

Equation 4.(公式 2)

在这里,Equation 5是最初的动力和Equation 6是最后的势头,与相同的约定时间用于Equation 7Equation 8。对象的作用力的总和等于物体的动量随时间变化。因此,如果对象没有净力,动量的变化将为零。说另一种方式,在一个封闭的系统与不受外力作用,最初的动力将等于最后的势头。

这一概念是最容易理解的一维、 二维碰撞。在一维碰撞,物体都有重量Equation 9和初始速度Equation 10碰撞与另一个物体都有重量Equation 11和初始速度Equation 12。在这些碰撞,将假定外部力量为太小了,有效果。在实验室中,气垫导轨用来减少摩擦,外力的作用,对滑翔飞行的鸟类。如果最初的动力是等于最后的势头,那么:

Equation 13(公式 3)

在那里引物的速度代表最后的流速与非底层的之间代表初始速度。

Figure 1
图 1.实验设置。

Procedure

1.理解光板计时器。

  1. 使用一种平衡,测量和记录每个滑翔机的质量。
  2. 与一个光板计时器在轨道上放置一个滑翔机。
  3. 设置为”门”设置的光板计时器。
  4. 当滑翔机通过光板时,它将记录上方滑翔机的旗穿过门的时间。在回程的时候,光板将不显示新的时间。开关切换来”读”,它将显示初始时间再加上第二次通过大门。
  5. 国旗是 10 厘米长;确定使用速度距离除以时间的滑翔机的速度。
  6. 发送通过光板滑翔机多次,包括回程后它已反弹远处的墙,和为了熟悉设备和测量速度。记住,速度有一个方向。让代表积极的初始速度方向和相反的方向,表示负速度值。

2.两个滑翔机的质量相等。

  1. 将两个滑翔机和两个光电计时器放在轨道上,如图 1所示。
  2. 使用方程 3确定最终速度的表达式。在此实验的一部分,滑翔机 B 将开始从休息。
  3. 给滑翔机一些初始的速度,所以它会碰撞滑翔机 B.记录 A,滑翔机的初始速度,以及每个滑翔机的最终速度。做了三次,记录您的结果,并比较了理论预测。

3.两个滑翔机的不同的质量。

  1. 添加 4 权滑翔机 B,其中将加倍它的质量。重复步骤 2.1-2.3。

4.平等群众不从休息

  1. 从滑翔机 B.删除权重
  2. 重复步骤 2.1-2.3,但给 B 的初始速度,以及滑翔机滑翔机 A.方向

动量守恒定律是物理学中最重要的法律之一,是经典力学中的许多现象的基础。

势头,通常由字母p,表示是产品的质量m和速度动量矩守恒原理指出,只要没有外力,作用应用对象变化的势头或 Δp,零。

相反,应用净的外部力量或F 净,一段时间结果动量的变化为该对象。动量守恒的现象也可以应用于对象的集合,这使得它对于学习物理碰撞非常有用。

本试验主要目的是动量的通过观察运动物体之间的碰撞测试守恒原理。

在深入的实验室实验之前, 让我们学习动量守恒的基本原则。牛顿定律是运动的理解动量矩守恒原理的核心。有关更多信息,请观看朱庇特的科学教育视频: 牛顿运动定律。

可以使用一个主球在台球桌上说明动量的概念。牛顿第二定律球杆净力赋予加速至质量为m的主球。加速度是速度v的变化在时间t。所以,如果我们将时间移到方程的另一边,我们是留下 Δmv或动量δ p的变化。因此,净力引起的动量的变化。

请注意,这个等式中的m是通常是恒定的所以动量的变化是依赖于速度在最后和初始参考点的差异。因为速度是一个矢量,正号或负号归因于其值,该值指示方向的运动。

在击打主球时的示例中,在 A 点 — — 在此方程 — — 由 vA 表示的初始速度为零。而在点 B 的最终速度是积极的。因此,动量的改变是由于棍子的净力积极。然后,当球从 B 点移到 C 点,假设有无外力作用在球上像摩擦或空气阻力, δ p将为零。

请注意,只可以动量守恒在孤立的系统-受净外力。

现在当主球移动从点 C 和罢工在点 D 表的一侧,其最终的速度变为零。因此,同时保留作为球遭到球杆同一数量级动量的改变变得负面。最后,当主球从墙上反弹,其最后到 E 点时速度负方向变动。我们知道,在点 D 的初始速度为零,因此动量的变化仍然是负因运动方向的改变。

这一现象的动量的改变和养护是用于研究碰撞,以及像两个游泳池球之间。请注意,在这种情况下两个球在一起会被视为孤立的系统。因此,机构的初始动量在碰撞前的总和等于之后他们最后的动量的总和。此外,一体的动量变化将平等和其他-反映了牛顿第三定律的截然相反。

请注意,这些池球碰撞将被视为弹性,意思动量和动能或柯的系统,保守的;但是这不是个案。其实更普遍遇到碰撞,如车祸、 都缺乏弹性和不可能遵守动量守恒,因为一些动能在碰撞过程中丢失。

现在,我们已检讨的动量守恒原则,让我们看看如何这些概念可以应用于涉及近无摩擦正轨滑翔机碰撞实验。

本实验主要包括平衡、 两个光板计时器、 两个滑翔机的质量相等、 额外重量、 空气供应、 气垫导轨与保险杠和一把尺子。

首先,用平衡测量群众的滑翔机,额外的重量,并记录这些值。接下来,连接空气供应到空气轨并打开它。气垫导轨用来减少摩擦,这将是一个外力对滑翔飞行的鸟类。

现在开始熟悉自己在轨道上放置一个滑翔机和组件之一的光板计时器计时过程。将定时器设置为 ‘门’ 设置和推动向光板滑翔机。上方的滑翔机的旗穿过光板时它将记录它的过境时间。知道的旗子 10 厘米长,将这个距离除以测量的时间得到的滑翔机速度。

滑翔机将反弹远保险杠和返回再次通过光板。光板显示初始的过境时间,并可以切换到 ‘读’ 设置显示返回的过境时间。重复该过程的测速滑翔机在初次和返回的行程,让自己熟悉的过程。因为速度是一个矢量,让积极的初始方向,返回方向为负值。

右边的第一套轨道上放置第二滑翔机和光板计时器。滑翔机 2 休息时,将推滑翔机 1,两个会发生碰撞。记录 1 滑翔机的初始速度以及每个滑翔机的最终速度。请注意,动量正在测量脉冲力系统是孤立的。三次重复此过程以得到多个数据集。

下一步,与在其原始位置滑翔机,地方加倍它的质量的一组额外的重量上滑翔机 2。对这大规模的配置重复前一组速度测量和记录这些值。

最后,将滑翔机重置为其原始位置,并从滑翔机 2 删除额外的重量。这套测量,滑翔机 2 将给初始速度这样两个滑翔机将接收推送前碰撞。记录的初始的和最终的速度,为每个滑翔机和三次重复此过程。

为第一个实验中涉及质量相等和滑翔机 1 最初移动,滑翔机 1 来到几乎完全停止后与滑翔机 2 发生碰撞。滑翔机 2 碰撞后的速度是类似于滑翔机 1 碰撞前的速度。因此,一个滑翔机动量的变化是平等和与其他的动量变化,这使它成为很好的例子,牛顿第三定律

不出所料,整个系统的初始的和最终的动量是几乎相等,反映动量守恒。在这些动量值差异都符合这种类型的实验包括测量误差和轨道不是完全水平的预期的错误。

第二个实验中涉及不平等群众,滑翔机 1 不来休息后碰撞与重的滑翔机,但反转方向后传授一些动力滑翔机 2。

再次,滑翔飞行的鸟类的动量变化是平等和相反而系统的总动量守恒的。近守恒系统动量,以及其初始的和最终的动能。这是因为碰撞是近弹性,因此可以忽略不计的外来磨擦力存在。

第三个实验中涉及的运动方向相反的质量相等的滑翔机,滑翔飞行的鸟类拥有相似的初始动量,然后同时保留他们的动量的大小相撞后扭转他们的方向。

系统总动量守恒在初始的和最终的动量值差异虽略大于先前的实验,由于所需的附加速度测量和摩擦可能更大损失。

动量守恒定律,虽然通常不被认为,原则是突出在各种各样的活动和事件。没有动量守恒火箭推进会不可能。最初,火箭,它的燃料都一动不动地和有动量为零。

然而,通过迅速驱逐了质量和动量的乏的燃料,被火箭弹向上,由于废弃燃料的相反方向的势头。这就解释了如何创建推力和推动空气或空间中没有对所有的事情推火箭。

火器的流量有一个显著的协会与动量守恒。

像火箭燃料系统,枪支弹药系统也开始在休息。当弹药发射出枪支以惊人的速度时,有反对对付它的势头。这被称为反冲,并且可以是非常强大的。

你刚看了动量守恒的朱庇特的简介。现在,您应该了解原理动量守恒和如何应用这了解的碰撞物理地解决问题。一如既往,感谢您收看 !

Results

表 1。从两个滑翔机的质量相等的结果。

滑翔机
(试行)
Equation 14
(cm/s)
Equation 15
(cm/s)
Equation 16
(cm/s)
Equation 17
(cm/s)
差异
(%)
(1) 72.5 -0.2
B (1) 0.0 67.1 72.5 66.9 8
A (2)。 35.6 0.3
B (2) 0.0 37.4 35.6 37.7 6
(3) 47.4 0.0
B (3) 0.0 47.8 47.4 47.8 1

表 2。结果从两个滑翔机的不同的质量。

滑翔机
(试行)
Equation 14
(cm/s)
Equation 15
(cm/s)
Equation 18
(公斤厘米/s)
Equation 19
(公斤厘米/s)
差异

(%)

(1) 52.9 -10.7
B (1) 0.0 37.7 52.9 64.7 22
A (2)。 60.2 -13.2
B (2) 0.0 41.5 60.2 69.8 16
(3) 66.2 -12.0
B (3) 0.0 45.9 66.2 79.7 20

表 3。从不从休息的质量相等的结果。

滑翔机
(试行)
Equation 14
(cm/s)
Equation 15
(cm/s)
Equation 16
(cm/s)
Equation 17
(cm/s)
差异
(%)
(1) 48.8 -29.9
B (1) -42.4 39.8 6.4 9.9 55
A (2)。 38.6 -25.2
B (2) -33.4 32.8 5.2 7.6 46
(3) 38.9 -43.1
B (3) -48.5 36.3 -9.6 -6.8 41

步骤 2、 3 和 4 的结果确认方程 3所作的预测。在步骤 2 中,滑翔机 A 来到后相撞滑翔机 B.几乎完全停止因此,几乎所有其动量传输到滑翔机 B.在步骤 3 中,滑翔机 A 不到站下车后与重滑翔机 B.相撞相反,它返回后传授一些动力滑翔机 B.向相反的方向在步骤 4 中,系统的总动量保持不变,尽管这两个滑翔机的方向变化。事实上,在某些情况下,总动量似乎会增加两个滑翔机下降的速度是与事实有关,那里是实验误差和碰撞本身并非完全弹性。声音和碰撞的废热可以采取系统的能量。这一事实,气垫导轨可能不是完全水平可以改变的速度滑翔飞行的鸟类的行为。如果甚至略微倾斜的轨道,速度会增加,由于重力的方向。结果还表明,系统,无论初始速度的总动量保持不变。

Applications and Summary

没有动量守恒,火箭将永远不会离开地面。火箭不实际上反对任何推-他们依靠推力起飞。最初,火箭和火箭本身的燃料都一动不动,和有动量为零。启动时,火箭推进了乏的燃料非常迅速。这乏的燃料的质量和动量。如果最后的势头必须等于最初的势头 (零),然后在相反的方向的废弃燃料必须有一些势头。因此,火箭是向上推进。

曾经开了一枪的人理解动量守恒。火箭的燃料系统从上面,像枪/弹药系统也开始在休息。当弹药发射出枪以惊人的速度时,在相反的方向来抵消子弹的势头必须有一些势头。这被称为反冲,并且可以是非常强大的。

由几个金属球,悬挂在字符串组成的流行桌装饰称为”牛顿摇篮”很好的理由。这是动量的另一个例子守恒。当一个球是解除和释放时,看来它的邻居,转移其势头。这一势头沿着线直到最后球第一,使其向外凸出的发展势头。这会走下去,如果不是因为外部力量,如空气阻力及能量损失的碰撞。

在这个实验中,通过考虑碰撞的两个滑翔机附近无摩擦的轨道上验证了动量守恒定律的法律。这个基本的法律也许是最重要的因为它能解决问题。如果有人知道初始动量,然后她知道最终的动量,反之亦然。

Transcript

Conservation of momentum is one of the most important laws in physics and underpins many phenomena in classical mechanics.

Momentum, typically denoted by the letter p, is the product of mass m and velocity v. The principle of momentum conservation states that an object’s change in momentum, or Δp, is zero provided no net external force is applied.

Conversely, applying a net external force, or F net, over a period of time results in a change in momentum for that object. The phenomenon of momentum conservation can also be applied to a collection of objects, which makes it useful for studying the physics of collisions.

The goal of this experiment is to test the principle of conservation of momentum by observing collisions between moving objects.

Before delving into the lab experiment, let’s study the basic principles of momentum conservation. Newton’s laws of motion are central to understanding the principle of momentum conservation. For more information, please watch JoVE’s Science Education video: Newton’s Laws of Motion.

The concepts of momentum can be illustrated using a cue ball on a pool table. Newton’s second law states that a net force applied by a cue stick imparts an acceleration a to a cue ball of mass m. Acceleration is the change in velocity v over time t. So, if we move time to the other side of the equation, we are left with Δmv, or the change in momentum Δp. Therefore, the net force gives rise to a change in momentum.

Note that the m in this equation is typically constant, so the change in momentum is dependent on the difference in velocities at the final and initial reference points. And since velocity is a vector quantity, a positive or negative sign is attributed to its value which indicates direction of motion.

In the cue ball example, the initial velocity at point A — denoted by vA in this equation — is zero. Whereas the final velocity at point B is positive. Thus, the momentum change is positive due to the net force applied by the stick. Then, when the ball is moving from point B to point C, assuming that there are no external forces acting on the ball like friction or air resistance, Δp would be zero.

Note that momentum can only be conserved in an isolated system – a system unaffected by net external forces.

Now, when the cue ball moves from point C and strikes the side of the table at point D, its final velocity becomes zero. Thus, the momentum change becomes negative while retaining the same magnitude as when the ball was struck by the cue stick. Lastly, when the cue ball rebounds off the wall, its final velocity at point E is negative due to change in direction. We know that the initial velocity at point D is zero, therefore the change in momentum remains negative because of the change in direction of movement.

This phenomenon of momentum change and conservation is useful for studying collisions as well, like between two pool balls. Note that in this case the two balls together would be treated as an isolated system. Therefore, the sum of the bodies’ initial momenta before the collision would equal the sum of their final momenta afterwards. Also, the momentum change of one body would be equal and opposite to that of the other – reflecting Newton’s third law.

Note that these pool ball collisions would be considered elastic, meaning that both momentum and kinetic energy or KE, of the system, are conserved; but this is not the case always. In fact, more commonly encountered collisions, such as car crashes, are inelastic and may not obey momentum conservation because some kinetic energy is lost during impact.

Now that we have reviewed the principles of momentum conservation, let’s see how these concepts can be applied to an experiment involving collisions of gliders on a near frictionless track.

This experiment consists of a balance, two photogate timers, two gliders of equal mass, additional weights, an air supply, an air track with bumpers, and a ruler.

First, using the balance, measure the masses of the gliders, the additional weights, and record these values. Next, connect the air supply to the air track and turn it on. An air track is used to reduce the amount of friction, which would be an external force on the gliders.

Now begin familiarizing yourself with the timing process by placing one glider and a component of one of the photogate timers on the track. Set the timer to the ‘gate’ setting and push the glider toward the photogate. When the flag above the glider passes through the photogate it will record its transit time. Knowing the flag is 10 centimeters long, divide this distance by the measured time to get the velocity of the glider.

The glider will bounce off the far bumper and return to pass through the photogate again. The photogate displays the initial transit time and can be switched to the ‘read’ setting to display the return transit time. Repeat the process of measuring the velocity of the glider during the initial and return trips to familiarize yourself with the process. Since velocity is a vector quantity, let the initial direction be positive and the return direction be negative.

Place a second glider and photogate timer on the track to the right of the first set. With glider 2 at rest, push glider 1 so that the two will collide. Record the initial velocity of glider 1 as well as the final velocities of each glider. Note that the momenta are being measured after the impulsive force has been applied and the system is isolated. Repeat this procedure three times to get multiple data sets.

Next, with the gliders in their original positions, place an additional set of weights on glider 2 that doubles its mass. Repeat the previous set of velocity measurements for this mass configuration and record these values.

Lastly, reset the gliders to their original positions and remove the additional weights from glider 2. For this set of measurements, glider 2 will be given an initial velocity such that both gliders will receive a push prior to the collision. Record the initial and final velocities for each glider and repeat this procedure three times.

For the first experiment involving equal masses and glider 1 initially moving, glider 1 comes to almost a complete stop after colliding with glider 2. And the velocity of glider 2 after collision is similar to the velocity of glider 1 before collision. Thus, the change in momentum of one glider is equal and opposite to the momentum change of the other, which makes this a good example of Newton’s 3rd Law

As expected, the initial and final momenta of the whole system are nearly equal, reflecting conservation of momentum. Discrepancies in these momenta values are consistent with errors expected for this type of experiment including measurement error and the track not being completely level.

For the second experiment involving unequal masses, glider 1 does not come to rest following the collision with the heavier glider, but reverses direction after imparting some momentum to glider 2.

Once again, the momentum changes of the gliders are equal and opposite while the momentum of the total system is conserved. The system momentum as well as its initial and final kinetic energies are nearly conserved. This is because the collision is nearly elastic and therefore negligible external friction forces are present.

For the third experiment involving gliders of equal mass moving in opposite directions, the gliders possess similar initial momenta and then reverse their directions after colliding while retaining their magnitudes of momenta.

The total system momentum is conserved although the discrepancies in the initial and final momentum values are slightly larger than the previous experiments owing to the additional velocity measurement required and potentially larger losses due to friction.

The principle of conservation of momentum, while not typically considered, is prominent in all manners of activities and events. Without momentum conservation rocket propulsion would not be possible. Initially the rocket and its fuel are motionless and have zero momentum.

However, by rapidly expelling spent fuel that has both mass and momentum, the rocket is propelled upward, as a result of the momentum in the opposite direction of the discarded fuel. This explains how rockets can create thrust and propel in air or space without pushing against anything.

The discharge of a firearm has a notable association with the conservation of momentum.

Like the rocket-fuel system, the firearm-ammunition system also starts at rest. When the ammunition is fired out of the firearm at a tremendous speed, there has to be opposing momentum to counter it. This is known as recoil and can be very powerful.

You’ve just watched JoVE’s introduction to Conservation of Momentum. You should now understand the principle momentum conservation and how this can be applied to solve problems and understand the physics of collisions. As always, thanks for watching!