Maximizing Transferred Energy between n bodies in linear collisions

In summary: But it's more likely that you'll wind up with a mess like this:In summary,The problem has two parts. In part 1, we have two point masses and there is another mass between them. They are all aligned in a line. Mass M is moving with speed u 1 toward mass M 1 and after collision all other masses are not moving. We want to find M such that the kinetic energy of M get maximum.In part 2, we have n masses between M and M 1 and 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
  • #1
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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  • #2
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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  • #3
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.
 
  • #4
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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1. How can the energy transfer between n bodies in linear collisions be maximized?

The energy transfer between n bodies in linear collisions can be maximized by ensuring that the collisions are perfectly elastic, meaning that there is no loss of energy due to deformation or friction. Additionally, the bodies should have equal masses and velocities, and the collisions should occur in a straight line.

2. What factors affect the energy transfer in linear collisions?

The energy transfer in linear collisions is affected by the masses and velocities of the bodies involved, as well as the angle and direction of the collisions. Other factors such as air resistance and surface friction can also impact the energy transfer.

3. How does the law of conservation of energy apply to linear collisions?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In linear collisions, the total kinetic energy of the bodies before the collision is equal to the total kinetic energy after the collision, as long as no external forces are present.

4. What is the difference between elastic and inelastic collisions?

In elastic collisions, the total kinetic energy of the bodies remains constant before and after the collision, while in inelastic collisions, some of the kinetic energy is lost due to deformation or other factors. Inelastic collisions are less efficient in terms of energy transfer compared to elastic collisions.

5. How can the efficiency of energy transfer in linear collisions be calculated?

The efficiency of energy transfer in linear collisions can be calculated by dividing the actual energy transferred by the maximum possible energy transfer, and then multiplying by 100 to get a percentage. This can be useful in determining the effectiveness of a collision system and identifying areas for improvement.

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