Car-Car System: Energy Conservation?

In summary, the conversation discusses the solution for a car-car system, which involves the law of conservation of energy and considers the velocities and masses of the cars. The thermal energy produced is found to be dependent on the frame of reference, but the kinetic energy post-collision is consistent in all frames. The conversation concludes that there is no need to ask further questions as the solution will be encountered again in the future. f
  • #1
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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
 
Last edited:
  • #2
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.
 
  • #3
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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