Maximizing Transferred Energy between n bodies in linear collisions

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SUMMARY

The discussion focuses on maximizing the transferred kinetic energy between multiple point masses during linear collisions. The first part establishes that for two masses, the optimal mass ##m_1## is determined to be ##\sqrt{mM}##, where ##M## is the moving mass and ##m## is the stationary mass. In the second part, the relationship for a series of n masses is proposed as each mass being the geometric mean of its adjacent masses, specifically ##m_i = \sqrt{m_{i-1}m_{i+1}}##. The participants also explore methods for formal proof, suggesting the use of mathematical induction or direct proof techniques.

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  • Understanding of elastic collisions and elasticity coefficients
  • Familiarity with kinetic energy concepts in physics
  • Knowledge of mathematical proofs, particularly induction and derivatives
  • Basic algebra and manipulation of equations involving multiple variables
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titansarus
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Homework Statement


My problem has two parts.
1) We have two point masses ##m,M##. and there is another mass ##m_1## between them.They are all aligned in a line. Mass ##M## is moving with speed ##u_1## toward ##m_1## and after collision and all other masses are not moving. we want to find ##m_1## such that the kinetic energy of ##m## get maximum.
2) Now think that we have n masses ##m_1 ,m_2 ,... m_n## between ##m , M##. find ##m_1 , m_2 , ... , m_n## such that the kinetic energy of ##m## get maximum. (elasticity coefficients are ##e_1## and ##e_2## ,... but they are not actually important and you can ignore them i.e. ##e=1##)

physics question 8.png


Homework Equations


In collision between two masses of mass ##m_1## and ##m_2## with elasticity coefficient ##e## moving with speed ##u_1## and ##u_2##, speed of ##v_1## after collision is:
## v_1 = \frac{(m_1-m_2 e)}{m_1+m_2} u_1 + \frac{m_2 e}{m_1 + m_2} u_2 ## (Eq.1)

The Attempt at a Solution


[/B]
The main goal is to maximize ##v## of ##m##. For part 1 if we write the (Eq.1) two times with ##u_1 = 0## for ##M,m_1## and ##m_1,m## we get:

##v_m = 4 ~e~e'~M u_1 (\frac{m_1}{(m+m_1)(M+m_1)})##. If we calculate ##\frac{d}{d~m_1} (v_m) =0## we get ##m_1 =\sqrt {mM}##.

For the second part, I don't know how to formally write a proof but I think the answer will be ##m_i = \sqrt{m_{i-1}m_{i+1}}##. which means that every mass must be the geometric mean of its left and right mass and for ##m_1## we get ##m_1 = \sqrt{M m_2}## and for ##m_n## we get ##m_n = \sqrt{m m_{n-1}}##. I think it is somehow obvious that we must maximize the ##v## in each collision but I can't mathematically prove that. So I want to know how to write a formal proof for this question.
 

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titansarus said:
So I want to know how to write a formal proof for this question.
Suppose you think you have somehow managed to tune all the intermediate masses so as to get the biggest final velocity you can. What if you find three adjacent masses in the sequence that do not fit the expected pattern. What can you deduce?
 
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haruspex said:
Suppose you think you have somehow managed to tune all the intermediate masses so as to get the biggest final velocity you can. What if you find three adjacent masses in the sequence that do not fit the expected pattern. What can you deduce?

Maybe we can say that if this thing happens for ##m_{i-1} , m_{i} , m_{i+1}## and they do not fit the expected pattern, then if we consider a ##v## for ##m_{i-1}##, then the energy (velocity) transferred to ##m_{i+1}## will not be maximum. However, I think maybe there should be a way to solve this without using "Proof by contradiction". Perhaps there is way using "Mathematical Induction" or even a direct proof like the derivative (max-min) proof for 3 mass as I said in the first post. But I cannot write it in a formal way for n-body.
 
titansarus said:
should be a way to solve this without using "Proof by contradiction".
It's a perfectly honourable style of proof.
titansarus said:
direct proof like the derivative (max-min) proof for 3 mass
That could get messy. You have n independent variables, so n differentiations to produce n simultaneous equations.
You could get lucky.
 
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