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I am currently studying Classical Mechanics, fifth edition, by Kibble and Berkshire. Problem 3 of chapter 1 is as follows:

Consider a system of three particles, each of mass $m$, whose motion is described by (1.9). If particles 2 and 3, even though not rigidly bound together, are regarded as forming a composite body of mass $2m$ located at the mid-point $\mathbf{r} = \dfrac{1}{2} (\mathbf{r}_2 + \mathbf{r}_3)$, find the equations describing the motion of the two-body system comprising particle 1 and the composite body (2+3). What is the force on the composite body due to particle 1? Show that the equations agree with (1.7). When the masses are unequal, what is the correct definition of the position of the composite (2+3) that will make (1.7) still hold?

I was unsure about this part:

When the masses are unequal, what is the correct definition of the position of the composite (2+3) that will make (1.7) still hold?

The answer is said to be

$$\mathbf{r} = \dfrac{m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3}{m_2 + m_3}.$$

(1.7) is as follows:

$$m_1 \mathbf{a}_1 = -m_2 \mathbf{a}_2$$

To try and understand how this could be done, I recently asked this question. Thanks to user Ja72's comments, I was able to do further research and learned that this is actually the center of mass:

https://en.wikipedia.org/wiki/Two-body_problem#Center_of_mass_motion_(1st_one-body_problem)

https://en.wikipedia.org/wiki/Center_of_mass#Barycentric_coordinates

http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html

The representations shown in these links reminded me of the law of conservation of momentum:

$$m_1\mathbf{v}_1 + m_2 \mathbf{v}_2 = m_1 \mathbf{v}_1^\prime + m_2 \mathbf{v}_2^\prime.$$

I then wondered: Does the law of conservation of momentum also hold for position and acceleration? Because, if it does, then it seems to me that we can represent the problem as follows:

$$m_1 \mathbf{r}_2 + m_2 \mathbf{r}_3 = (m_1 + m_2) \mathbf{r}^\prime \\ \Rightarrow \mathbf{r}^\prime = \dfrac{m_1 \mathbf{r}_2 + m_2 \mathbf{r}_3}{m_1 + m_2},$$

where $(m_1 + m_2)$ is the composite mass. This seems to be of the form that we are looking for. Furthermore, unlike user Ja72's answer here, it uses basic equations of classical mechanics that are discussed in chapter 1, which means that it is more likely to be the solution method that the authors were intending the reader to use to solve this problem.

I had the question of whether the law of conservation of momentum also holds for position and acceleration answered here by the user Dale. They said that conservation of position is not valid, and that, in general, conservation laws do not hold whenever the center of mass of the system is moving. So how do I reconcile this with the fact that using the conservation laws in terms of position seems to get us the correct solution of $\mathbf{r} = \dfrac{m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3}{m_2 + m_3}$? Is it because the center of mass is actually not moving in this case? Or is it just a coincidence? I would greatly appreciate it if people would please take the time to clarify this.

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Conservation of position is not invalid, it's meaningless. To get the velocity and acceleration of the center of mass, take the first and second derivative of the formula for the position. In the absence of external forces, the momentum is conserved in any inertial frame of reference. (The velocity of the center of mass is different but constant in each.) In a collision, energy is conserved only if the collision is fully elastic.

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  • $\begingroup$ Thanks for the answer. So what is your assessment with regards to $\mathbf{r} = \dfrac{m_2 \mathbf{r}_2 + m_3 \mathbf{r}_3}{m_2 + m_3}$? $\endgroup$ – The Pointer Jun 13 '20 at 12:28
  • $\begingroup$ That locates the center of mass for those two masses. $\endgroup$ – R.W. Bird Jun 13 '20 at 17:57
  • $\begingroup$ Yes, but my concern is its derivation from the force laws of classical mechanics (see the problem statement). $\endgroup$ – The Pointer Jun 13 '20 at 17:58
  • $\begingroup$ Keeping in mind that that formula can include any number of masses, the first derivative tells you that the momentum of the center of mass is equal to the vector sum of the momenta of all of the separate masses. The second derivative tells you that the vector sum of all the forces acting on the separate masses determines the acceleration of the center of mass. (Newton' s second law). Either of these can be taken as the definition of the center of mass. $\endgroup$ – R.W. Bird Jun 13 '20 at 18:33
  • $\begingroup$ Yes, but the problem statement refers to the definition of the position of the composite (which, in this case, is the definition of the position of the center of mass). $\endgroup$ – The Pointer Jun 13 '20 at 18:39

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