Car-Car System: Energy Conservation?

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The discussion revolves around the feasibility of a car-car energy conservation system, emphasizing the application of energy conservation laws. It is established that the thermal energy produced during a collision can be calculated as mv², regardless of the reference frame used. The calculations confirm that the energy before and after the collision remains consistent across different frames. The importance of considering all energies involved in the system is highlighted, although the initial inquiry about reassurance indicates some uncertainty. Overall, the analysis provides clarity on the energy dynamics in a car-car system.
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Homework Statement
(I) Two railroad cars, each of mass 56,000 kg, are traveling
##95km/h##toward each other. They collide head-on and come
to rest. How much thermal energy is produced in this collision?
Relevant Equations
law of conserfvation of energy
this is an easy problem but would it be possible to consider car-car system. What I did on paper was carsystem and because they have the same properties(mass en speed) multiply by ##2##

solution for car-car-earth system I assume is the following if it is possible?

solution for car-car:
law or conservation of energy says:

##E_1 = E_2 + W_{th}##
##\frac12m_{c1}v_{1,c1}^2 + \frac12m_{c2}v_{2,c2}^2 = 0 + W_{th}##
massses same, velocity same so :
##W_{th} = mv^2 = 3.9E7J##

edit: What I actually want reassurance of is.. this is all dependent on the system and if we take car-car system we consider all the energies of the involved objects by examining all of 'em seperately. I am asking stuff like this because it's kinda important but not really mentioned only implicitly in the book
Thanks in advance

Edit: ;earth not included
 
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It is not dependent on the frame. If the cars have velocities ##\pm v## in the cener of mass frame then the thermal energy produced is indeed ##mv^2## as you have concluded. If you instead consider a frame moving with velocity ##u## relative to the center of mass system then the velocities of the cars are ##\pm v - u## and consequently the energy before collision
$$
W_0 = \frac{m}{2}[(v-u)^2 +(v+u)^2] = m(v^2 + u^2).
$$
After collision the cars are moving at velocity ##u## and so the kinetic energy post-collision is
$$
W_k = \frac{2m}{2} u^2 = m u^2.
$$
The thermal energy produced is therefore ##W_0-W_k = mv^2##, just as found in the center of mass frame.
 
Orodruin said:
It is not dependent on the frame. If the cars have velocities ##\pm v## in the cener of mass frame then the thermal energy produced is indeed ##mv^2## as you have concluded. If you instead consider a frame moving with velocity ##u## relative to the center of mass system then the velocities of the cars are ##\pm v - u## and consequently the energy before collision
$$
W_0 = \frac{m}{2}[(v-u)^2 +(v+u)^2] = m(v^2 + u^2).
$$
After collision the cars are moving at velocity ##u## and so the kinetic energy post-collision is
$$
W_k = \frac{2m}{2} u^2 = m u^2.
$$
The thermal energy produced is therefore ##W_0-W_k = mv^2##, just as found in the center of mass frame.
wow that's aamazinggg thanks a lot! I guess I don't need to ask such question then as I'll most likely encounter that soon myself
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...